Calculus 1 Exam 4 Review Sheet

Problem 1

Find 3 different antiderivatives of (x^3).

Solution

We know that if f(x) is one antiderivative, then all antiderivatives have the form f(x)+c where c is a number.  (d/(dx))(((x^4)/4))=(x^3), so ((x^4)/4) is one answer (obtained via “raise the power one, divide by the new power”).  Randomly picking two constants, ((x^4)/4)+17 and ((x^4)/4)-987 are two others.

Problem 2

Find an antiderivative for 3sin(x)-4x+(x^5).

Solution

We know that when taking a derivative we can split it up across addition/subtraction and that constant multiples bcan be pulled out front.   So we can just focus on finding antiderivatives for sin(x), x, and (x^5).  (d/(dx))cos(x)=-sin(x) which is off by a minus sign, so an antiderivative for sin(x) is -cos(x).  Using the “raise the power one, divide by the new power” idea for x=(x^1) and (x^5), antiderivatives for those are ((x^2)/2) and ((x^6)/6).  So an antiderivative for 3 sin(x)-4x+(x^5) is 3(-cos(x)) -4 ((x^2)/2)+((x^6)/6)=-3cos(x)-2(x^2)+((x^6)/6).

Note this problem (as well as problem 1) is a precursor to the “find all antiderivatives”/“find the indefinite integral” type of problem where we tend to use the established rules rather than going back to basics every time.

Problem 3

Find ∫4(sec^2)(x)-(Sqrt[x])dx.

Solution

Since we will need to use the “power rule for integrals”, it’s best to convert (Sqrt[x]) to (x^(1/2)) before integrating.  Once you’ve done that ∫4(sec^2)(x)-(x^(1/2))dx=4∫(sec^2)(x)dx-∫(x^(1/2))dx=4tan(x)-((x^(3/2))/(3/2))+C=4tan(x)-(2/3)(x^(3/2))+C.

Problem 4

Find ∫(5/(x^9))-(3/((sin^2)(x)))-8(x^9)dx

Solution

In terms of integrating it will be easier if we convert (5/(x^9)) to 5(x^(-9))so we can use the power rule.  In addition we don’t have a familiar rule for integrating (1/((sin^2)(x))); but using trigonometry we know (1/((sin^2)(x)))=(csc^2)(x) and we do have a rule for that.  So ∫(5/(x^9))-(3/((sin^2)(x)))-8(x^9)dx=∫5(x^(-9))-3(csc^2)(x)-8(x^9)dx=5∫(x^(-9))dx-3∫(csc^2)(x)dx-8∫(x^9)dx=5((x^(-8))/(-8))-3(-cot(x))-8((x^10)/10)+C=-(5/(x^8))+3cot(x)-(4/5)(x^10)+C

Problem 5

Find ∫(1/(x^(1/4)))+2sec(x)tan(x)dx

Solution

Rewriting (1/(x^(1/4)))=(x^(-1/4)), ∫(x^(-1/4))+2sec(x)tan(x)dx=∫(x^(-1/4))+2∫sec(x)tan(x)dx=((x^(3/4))/(3/4))+2sec(x)+C=(4/3)(x^(3/4))+2sec(x)+C

Problem 6

Given that a falling object’s acceleration is given by a(t)=-32 ft/(sec^2) and that at time 0 its velocity is 40 feet/sec upward and its height is 5 feet, find its position at any time.  When does it land?

Solution

Let s(t) represent the position of the ball at time t (with positive values meaning upward), v(t) the velocity at any time t, and a(t)=-32 feet/(sec^2) the acceleration at time t.  We know that s'(t)=v(t) and v'(t)=a(t); rethinking these equations in terms of antiderivatives means velocity is an antiderivative for acceleration and position is an antderivative for velocity.  Dealing with these in two phases:

Finding the velocity formula:

	We know that v(t)=∫a(t)dt=∫-32dt=-32t+(Subscript[c, 1]), where (Subscript[c, 1]) is a constant.  We also know that v(0)=40 feet/sec.  Therefore v(0)=40=-32(0)+(Subscript[c, 1]), or (Subscript[c, 1])=40.  Therefore the velocity formula at any time (until it lands anyway) is given by v(t)=-32t+40.
	
Finding the position formula:

	We know that s(t)=∫v(t)dt=∫-32t+40dt=-32×(1/2)(t^2)+40t+(Subscript[c, 2])=-16(t^2)+40t+(Subscript[c, 2]), where (Subscript[c, 2]) is some unknown constant.  We also know that s(0)=5 feet.  Putting these together gives s(0)=5=-16×(0^2)+40×0+(Subscript[c, 2])=(Subscript[c, 2]), so (Subscript[c, 2])=5.  Therefore the formula for the height of the ball at any time t is s(t)=-16(t^2)+40t+5.
	
The landing time:
We want to know at what time the ball will land: that is for what t will s(t)=0?  As we have the formula for the height this becomes -16(t^2)+40t+5=0; applying the quadratic formula yields t=((-40±(Sqrt[(40^2)-4(-16)5]))/(2×(-16)))=((-40±(Sqrt[1600+320]))/(-32))=((-40±(Sqrt[1920]))/(-32))=((-40±8(Sqrt[30]))/(-32))=((-5±(Sqrt[30]))/(-4))=(5/4)±((Sqrt[30])/4).  As the time would have to be positive, the landing time must be (5/4)+((Sqrt[30])/4)≈2.619 seconds.

Problem 7

When do two functions have the same derivative?

Solution

If f(x) and g(x) are two functions whose derivatives match on an interval I, then there must be a constant c with f(x)=g(x)+c on the interval I.  If the range of numbers is not an interval but the union of different intervals, then you must have a constant c for each interval with f(x)=g(x)+c (in other words, f(x) is g(x) plus a piecewise constant function).

For an example of this more complicated situation, consider the function -(1/(x^2)).  This is defined in the two non-connected pieces x>0 and x<0.  Using the power rule gives us that -((x^(-1))/(-1))=(1/x) is an antiderivative.  But I can make a new antiderivative from this by choosing a one constant to add when x>0 and another constant when x<0; for example another antiderivative could be f(x)=({{{, ({{(1/x)+5, when x>0}, {(1/x)-9, when x<0}})}})  .  The derivative of f(x) is still -(1/(x^2)), but there is no single constant c with (1/x)=f(x)+c.

Problem 8

Break down (Underoverscript[∑ , (k=1), 20])(2(Subscript[a, k])-3(Subscript[b, k])+4) into simpler sums.

Solution

We know summation notation breaks up across + signs and - signs and that constant multiples can be pulled out front; so (Underoverscript[∑ , (k=1), 20])2(Subscript[a, k])-3(Subscript[b, k])+4=2(Underoverscript[∑ , (k=1), 20])(Subscript[a, k])-3(Underoverscript[ ∑, (k=1), 20])(Subscript[b, k])+(Underoverscript[ ∑, (k=1), 20])4.  The last sum is equal to 4×20=80, so we have (Underoverscript[ ∑, (k=1), 20])2(Subscript[a, k])-3(Subscript[b, k])+4=2(Underoverscript[ ∑, (k=1), 20])(Subscript[a, k])-3(Underoverscript[ ∑, (k=1), 20])(Subscript[b, k])+80.

Problem 9

Write out (Underoverscript[ ∑, (k=2), 11])(1/(Sqrt[k])) without using summation notation.

Solution

(Underoverscript[ ∑, (k=2), 11])(1/(Sqrt[k]))=(1/(Sqrt[2]))+(1/(Sqrt[3]))+(1/(Sqrt[4]))+(1/(Sqrt[5]))+(1/(Sqrt[6]))+(1/(Sqrt[7]))+(1/(Sqrt[8]))+(1/(Sqrt[9]))+(1/(Sqrt[10]))+(1/(Sqrt[11]))

Problem 10

Write the sum 2+5+8+11+...+152+155 using sigma notation and use our summation formula to find the value of the sum.

Solution

If you look at the sum 2+5+8+11+....+152+155 you’ll notice that the numbers go up by 3 each time.  If we try to use the formula 3k to match the pattern, we would get the numbers 3, 6, 9, 12... which are all 1 too high.  So instead if we use the formula 3k-1 we would get 2, 5, 8, 11, 14 and so on.  Note that k=1 corresponds to the number 2 and k=52 corresponds to the final number 155.  So we can represent 2+5+8+...+152+155 as (Underoverscript[∑ , (k=1), 52])3k-1.  Note that the formula 3k+2 also fits the same string of numbers, but would need to use k=0 to correspond to the first 2 and we prefer our sums to start with k=1 where possible (at least for now).

(Underoverscript[ ∑, (k=1), 52])3k-1=3(Underoverscript[ ∑, (k=1), 52])k-(Underoverscript[ ∑, (k=1), 52])1.  We know that (Underoverscript[ ∑, (k=1), n])k=((n(n+1))/2), so (Underoverscript[ ∑, (k=1), 52])k=((52×53)/2)=1378. (Underoverscript[∑ , (k=1), 52])1=52.  So (Underoverscript[ ∑, (k=1), 52])3k-1=3(Underoverscript[ ∑, (k=1), 52])k-(Underoverscript[ ∑, (k=1), 52])1=3×1378-52=4082.

Problem 11

State the summation formulae for (Underoverscript[∑ , (k=1), n])1, (Underoverscript[∑ , (k=1), n])k, (Underoverscript[∑ , (k=1), n])(k^2), and (Underoverscript[∑ , (k=1), n])(k^3).

Solution

(Underoverscript[ ∑, (k=1), n])1=n
(Underoverscript[ ∑, (k=1), n])k=((n(n+1))/2)
(Underoverscript[ ∑, (k=1), n])(k^2)=((n(n+1)(2n+1))/6)
(Underoverscript[ ∑, (k=1), n])(k^3)=((((n^2)(n+1))^2)/4)
Note that in general (Underoverscript[ ∑, (k=1), n])(k^j) will be a polynomial in n of degree j+1.

Problem 12

Find (Underoverscript[ ∑, (k=1), 100])(3+2k+5(k^2))

Solution

(Underoverscript[ ∑, (k=1), 100])3+2k+5(k^2)=(Underoverscript[ ∑, (k=1), 100])3+2(Underoverscript[ ∑, (k=1), 100])k+5(Underoverscript[ ∑, (k=1), 100])(k^2)=3×100 + 2×((100(101))/2)+5×((100(101)(201))/6)=300+10100+1691750=1702150.

Problem 13

Find (2.01^3)+(2.02^3)+.... + (5^3)

Solution

First we need to rewrite this sum using summation notation.  Note that the numbers being cubed are going up by .01=(1/100) starting at 2.01 and ending at 5.  This can be modeled in the formula 2+(k/100) as k goes from 1 to 300.  Each of those numbers is then cubed before adding them up.  So we can write the sum as (Underoverscript[∑ , (k=1), 300])(((2+(k/100)))^3).  From the “1 3 3 1” row of Pascal’s triangle we know that (((2+(k/100)))^3)=(2^3)+3×(2^2)×(k/100)+3×2×((((k/100)))^2)+((((k/100)))^3)=8+((3k)/25)+((3(k^2))/5000)+((k^3)/1000000).  So  (Underoverscript[∑ , (k=1), 300])(((2+(k/100)))^3)=(Underoverscript[ ∑, (k=1), 300])(8+((3k)/25)+((3(k^2))/5000)+((k^3)/1000000))=(Underoverscript[ ∑, (k=1), 300])8+(3/25)(Underoverscript[ ∑, (k=1), 300])k+(3/5000)(Underoverscript[ ∑, (k=1), 300])(k^2)+(1/1000000)(Underoverscript[ ∑, (k=1), 300])(k^3)=300×8+(3/25)×((300(301))/2)+(3/5000)×((300(301)(601))/6)+(1/1000000)×(((300^2)(301^2))/4)=(6113421/400)=15283.5525

Problem 14

Use Riemann sums with 10 equal subintervals and right-hand endpoints to estimate the area under y=(1/x) from x=1 to x=2.  Repeat using left-hand endpoints.  How do these compare to the actual endpoints?

Solution

For right-hand endpoints:

As the interval [1,2] has a length of 1 and we are using a uniform partition of 10 pieces, the length of each piece is given by Δ (Subscript[x, k])=Δ x=(1/10).  The formula for the right-hand endpoint of the k-th piece is given by (Subscript[c, k])=1+k Δ x=1+k(1/10)=1+(k/10).  So our Riemann sum for f(x)=(1/x) in this case is given by R=(Underoverscript[ ∑, (k=1), 10])f((Subscript[c, k]))Δ (Subscript[x, k])=(Underoverscript[ ∑, (k=1), 10])f(1+(k/10))×(1/10)=(Underoverscript[ ∑, (k=1), 10])(1/(1+(k/10)))×(1/10)=(Underoverscript[ ∑, (k=1), 10])(1/(10+k)).  We don’t have a formula to help us with this sum, so we have to write it longhand as (1/11)+(1/12)+(1/13)+...+(1/20)≈0.6687714032.  The picture for this Riemann sum looks like this:

So it appears that the right-hand Riemann sum is the lower sum, which is an underestimate (this is true because (1/x) is decreasing).

For left-hand endpoints:

As the interval [1,2] has a length of 1 and we are using a uniform partition of 10 pieces, the length of each piece is given by Δ (Subscript[x, k])=Δ x=(1/10).  The formula for the left-hand endpoint of the k-th piece is given by (Subscript[c, k])=1+(k-1) Δ x=1+(k-1)(1/10)=1+((k-1)/10).  So our Riemann sum for f(x)=(1/x) in this case is given by R=(Underoverscript[ ∑, (k=1), 10])f((Subscript[c, k]))Δ (Subscript[x, k])=(Underoverscript[ ∑, (k=1), 10])f(1+((k-1)/10))×(1/10)=(Underoverscript[ ∑, (k=1), 10])(1/(1+((k-1)/10)))×(1/10)=(Underoverscript[ ∑, (k=1), 10])(1/(9+k)).  We don’t have a formula to help us with this sum, so we have to write it longhand as (1/10)+(1/12)+(1/13)+...+(1/19)≈0.71877140317542794322980081494013382558`10..  The picture for this Riemann sum looks like this:

The left-hand sum turns out to be the upper sum (which as before makes sense as the curve is decreasing), the the left-hand sum is an overestimate.

This tells us the true area A lies somewhere between these two values, so 0.6687714032<A<0.71877140317542794322980081494013382558`10..  As it turns out the true value for the sum is ln(2)≈0.69314718055994530937448036556070007919`10..

Problem 15

Use Riemann sums with 10 equal intervals and right-hand endpoints to estimate the area under y=(Sqrt[x]) from x=0 to x=4.  Repeat Using left-hand endpoints.  How do these compare to the actual area under the curve?

Solution

For right-hand endpoints:

As the interval [0,4] has a length of 4 and we are using a uniform partition of 10 pieces, the length of each piece is given by Δ (Subscript[x, k])=Δ x=(4/10)=(2/5).  The formula for the right-hand endpoint of the k-th piece is given by (Subscript[c, k])=0+k Δ x=0+k(2/5)=((2k)/5).  So our Riemann sum for f(x)=(1/x) in this case is given by R=(Underoverscript[ ∑, (k=1), 10])f((Subscript[c, k]))Δ (Subscript[x, k])=(Underoverscript[ ∑, (k=1), 10])f(((2k)/5))×(2/5)=(Underoverscript[ ∑, (k=1), 10])(Sqrt[((2k)/5)])×(2/5)=(Underoverscript[ ∑, (k=1), 10])((2(Sqrt[2]))/(5(Sqrt[5])))(Sqrt[k])=((2(Sqrt[2]))/(5(Sqrt[5])))((Sqrt[1])+(Sqrt[2])+(Sqrt[3])+...+(Sqrt[10]))≈5.68407

The picture for this Riemann sum looks like this:

We can see this is the same as an upper sum (which makes sense as (Sqrt[x]) is increasing) so this is an overestimate.

For left-hand endpoints:

As the interval [0,4] has a length of 4 and we are using a uniform partition of 10 pieces, the length of each piece is given by Δ (Subscript[x, k])=Δ x=(4/10)=(2/5).  The formula for the left-hand endpoint of the k-th piece is given by (Subscript[c, k])=0+(k-1) Δ x=0+(k-1)(2/5)=((2(k-1))/5).  So our Riemann sum for f(x)=(1/x) in this case is given by (Underoverscript[ ∑, (k=1), 10])f((Subscript[c, k]))Δ (Subscript[x, k])=(Underoverscript[ ∑, (k=1), 10])f(((2(k-1))/5))×(2/5)=(Underoverscript[ ∑, (k=1), 10])(Sqrt[((2(k-1))/5)])×(2/5)=(Underoverscript[ ∑, (k=1), 10])((2(Sqrt[2]))/(5(Sqrt[5])))(Sqrt[k-1])=((2(Sqrt[2]))/(5(Sqrt[5])))((Sqrt[0])+(Sqrt[1])+(Sqrt[2])+...+(Sqrt[9]))≈4.88407473365454`

The picture for this Riemann sum looks like this:

This is the same as the lower sum (which again makes good sense) so this must be an underestimate.  Therefore the true value A must lie between these two estimates; that is 4.88407<A<5.68407.

Problem 16

Find  (Subscript[∫ , 0]^2)(x^2)dx using Riemann sums.

Solution

Let (Subscript[R, n]) represent the Riemann sum for f(x)=(x^2) on [0,2] which uses right-hand endpoints and n equal pieces.  As (x^2) is continuous on [0,2], (Subscript[∫ , 0]^2)(x^2)dx=(Underscript[ lim, n⟶∞]) (Subscript[R, n]).  As the interval has a length of 2 and we are splitting it into n equal pieces, the length of each piece is given by Δ (Subscript[x, k])=Δ x=(2/n).  In addition the right-hand endpoint of the k-thpiece is given by (Subscript[c, k])=0+k Δ x=k×(2/n)=((2k)/n).  Therefore the formula for the general Riemann sum is (Subscript[R, n])=(Underoverscript[ ∑, (k=1), n])f((Subscript[c, k]))Δ (Subscript[x, k])=(Underoverscript[ ∑, (k=1), n])f(((2k)/n))×(2/n)=(Underoverscript[ ∑, (k=1), n])(((((2k)/n)))^2)×(2/n)=(Underoverscript[ ∑, (k=1), n])((8(k^2))/(n^3))=(8/(n^3))(Underoverscript[ ∑, (k=1), n])(k^2)=(8/(n^3))((n(n+1)(2n+1))/6)=((4(n+1)(2n+1))/(3(n^2)))

Now (Subscript[∫ , 0]^2)(x^2)dx=(Underscript[lim , n⟶∞]) (Subscript[R, n])=(Underscript[ lim, n⟶∞])((4(n+1)(2n+1))/(3(n^2)))=(Underscript[ lim, n⟶∞]) ((4(1+(1/n))(2+(1/n)))/3)=((4(1)(2))/3)=(8/3).

Note that as (x^2)≥0 this is a true are in the geometric sense:

Plot[ x^2, {x,0,2}, Filling-> Axis, PlotLabel-> TraditionalForm[y==x^2]]

Problem 17

Find (∫_( 1)^4) 2-(x^3)dx using Riemann sums.

Solution

Let (Subscript[R, n]) represent the Riemann sum for f(x)=2-(x^3) on [1,4] which uses right-hand endpoints and n equal pieces.  As 2-(x^3) is continuous on [1,4], (∫_( 1)^4) 2-(x^3)dx=(Underscript[ lim, n⟶∞]) (Subscript[R, n]).  As the interval has a length of 3 and we are splitting it into n equal pieces, the length of each piece is given by Δ (Subscript[x, k])=Δ x=(3/n).  In addition the right-hand endpoint of the k-thpiece is given by (Subscript[c, k])=1+k Δ x=1+k×(3/n)=1+((3k)/n).  Therefore the formula for the general Riemann sum is 
(Subscript[R, n])=(Underoverscript[ ∑, (k=1), n])f((Subscript[c, k]))Δ (Subscript[x, k])=(Underoverscript[ ∑, (k=1), n])f(1+((3k)/n))×(3/n)=(Underoverscript[ ∑, (k=1), n]) (2-(((1+((3k)/n)))^3))×(2/n)=(3/n)(Underoverscript[ ∑, (k=1), n]) (2-(1+((9k)/n)+((27(k^2))/(n^2))+((27(k^3))/(n^3)))=(3/n)(Underoverscript[ ∑, (k=1), n]) (1-((9k)/n)-((27(k^2))/(n^2))-((27(k^3))/(n^3)))=(3/n)((Underoverscript[∑ , (k=1), n])1-(9/n)(Underoverscript[∑ , (k=1), n])k-(27/(n^2))(Underoverscript[∑ , (k=1), n])(k^2)-(27/(n^3))(Underoverscript[∑ , (k=1), n])(k^3))=(3/n)(n-(9/n)((n(n+1))/2)-(27/(n^2))((n(n+1)(2n+1))/6)-(27/(n^3))((((n^2)(n+1))^2)/4))=3-((27(n+1))/(2n))-((27(n+1)(2n+1))/(2(n^2)))-((81(((n+1))^2))/(4(n^2)))

Now (∫_( 1)^4) 2-(x^3)dx=(Underscript[ lim, n⟶∞]) (Subscript[R, n])=(Underscript[ lim, n⟶∞])3-((27(n+1))/(2n))-((27(n+1)(2n+1))/(2(n^2)))-((81(((n+1))^2))/(4(n^2)))=(Underscript[ lim, n⟶∞])3-(27/2)(1+(1/n))-(27/2)(1+(1/n))(2+(1/n))-(81/4)(((1+(1/n)))^2)=3-(27/2)-(54/2)-(81/4)=-(231/4)

Note that this indicates the vast majority of the area lies below the x-axis, as seen in the picture below:

Plot[ 2-x^3, {x,1,4}, PlotLabel-> y==2-x^3, Filling-> Axis, FillingStyle-> {LightRed, Orange}, ImageSize-> 300]

Problem 18

Find (∫_( 0)^( 6)) 3-x+(x^2)dx using Riemann sums

Solution

Let (Subscript[R, n]) represent the Riemann sum for f(x)=3-x+(x^2) on [0,6] which uses right-hand endpoints and n equal pieces.  As 3-x+(x^2) is continuous on [0,2], (∫_( 0)^( 6)) 3-x+(x^2)dx=(Underscript[ lim, n⟶∞]) (Subscript[R, n]).  As the interval has a length of 6 and we are splitting it into n equal pieces, the length of each piece is given by Δ (Subscript[x, k])=Δ x=(6/n).  In addition the right-hand endpoint of the k-thpiece is given by (Subscript[c, k])=0+k Δ x=k×(6/n)=((6k)/n).  Therefore the formula for the general Riemann sum is (Subscript[R, n])=(Underoverscript[ ∑, (k=1), n])f((Subscript[c, k]))Δ (Subscript[x, k])=(Underoverscript[ ∑, (k=1), n])f(((6k)/n))×(6/n)=(6/n)(Underoverscript[ ∑, (k=1), n])(3-((6k)/n)+(((((6k)/n)))^2))=(6/n)(Underoverscript[ ∑, (k=1), n])(3-((6k)/n)+((36(k^2))/(n^2)))=(6/n)((Underoverscript[ ∑, (k=1), n])3-(6/n)(Underoverscript[ ∑, (k=1), n])k+(36/((n^2) ))(Underoverscript[ ∑, (k=1), n])(k^2))=(6/n)(3n-(6/n)((n(n+1))/2)+(36/(n^2))((n(n+1)(2n+1))/6))=18-((18(n+1))/n)+((36(n+1)(2n+1))/(n^2)).
Now (∫_( 0)^( 6)) 3-x+(x^2)dx=(Underscript[lim , n⟶∞]) (Subscript[R, n])=(Underscript[ lim, n⟶∞])18-((18(n+1))/n)+((36(n+1)(2n+1))/(n^2))=(Underscript[ lim, n⟶∞]) 18-18(1+(1/n))+36(1+(1/n))(2+(1/n))=18-18+72=72

Note that as 3-x+(x^2)≥0, this is a true area in the geometric sense:

Plot[ 3-x+x^2, {x,0,6}, PlotLabel-> TraditionalForm[ y==3-x+x^2], Filling->Axis, ImageSize->300]

Problem 19

Given that (Underoverscript[ ∑, (k=1), n])(k^4)=((n(n+1)(2n+1)(3(n^2)+3n-1))/30), find (∫_(  1)^( 2))(x^4)dx

Solution

Let (Subscript[R, n]) represent the Riemann sum for f(x)=(x^4) on [1,2] which uses right-hand endpoints and n equal pieces.  As (x^4) is continuous on [1,2], (∫_(  1)^( 2))(x^4)dx=(Underscript[ lim, n⟶∞]) (Subscript[R, n]).  As the interval has a length of 1 and we are splitting it into n equal pieces, the length of each piece is given by Δ (Subscript[x, k])=Δ x=(1/n).  In addition the right-hand endpoint of the k-thpiece is given by (Subscript[c, k])=1+k Δ x=1+k×(1/n)=1+(k/n).  Therefore the formula for the general Riemann sum is (Subscript[R, n])=(Underoverscript[ ∑, (k=1), n])f((Subscript[c, k]))Δ (Subscript[x, k])=(Underoverscript[ ∑, (k=1), n])f(1+(k/n))×(1/n)=(1/n)(Underoverscript[ ∑, (k=1), n])(((1+(k/n)))^4)=(1/n)(Underoverscript[ ∑, (k=1), n])(1+((4k)/n)+((6(k^2))/(n^2))+((4(k^3))/(n^3))+((k^4)/(n^4)))=(1/n)((Underoverscript[ ∑, (k=1), n])1+(4/n)(Underoverscript[ ∑, (k=1), n])k+(6/(n^2))(Underoverscript[ ∑, (k=1), n])(k^2)+(4/(n^3))(Underoverscript[ ∑, (k=1), n])(k^3)+(1/(n^4))(Underoverscript[ ∑, (k=1), n])(k^4))
=(1/n)(n+(4/n)((n(n+1))/2)+(6/(n^2))((n(n+1)(2n+1))/6)+(4/(n^3))((((n^2)(n+1))^2)/4)+(1/(n^4))((n(n+1)(2n+1)(3(n^2)+3n-1))/30))=1+((2(n+1))/n)+(((n+1)(2n+1))/(n^2))+((((n+1))^2)/(n^2))+(((n+1)(2n+1)(3(n^2)+3n-1))/(30(n^4)))

Now (∫_(  1)^( 2))(x^4)dx=(Underscript[ lim, n⟶∞]) (Subscript[R, n])=(Underscript[ lim, n⟶∞]) 1+((2(n+1))/n)+(((n+1)(2n+1))/(n^2))+((((n+1))^2)/(n^2))+(((n+1)(2n+1)(3(n^2)+3n-1))/(30(n^4)))=(Underscript[ lim, n⟶∞])1+2(1+(1/n))+(1+(1/n))(2+(1/n))+(((1+(1/n)))^2)+(((1+(1/n))(2+(1/n))(3+(3/n)-(1/(n^2))))/30)=1+2+2+1+(1/5)=(31/5).

Note that as (x^4)≥0 on [1,2] this is a true geometric area:

Plot[ x^4, {x,1,2}, PlotLabel-> y==x^4, Filling->Axis]

Problem 20

Explain the difference between net/signed area and total area.

Solution

Total area is the idea of geometric area; all area is counted as positive.  In net/signed area any area that is below the x-axis counts as negative, so the complete area under a curve may be positive, negative or zero.  This will be important for many applications as the definite integral (which is net area) can represent far more than area - it can be net work, net worth, etc. - and these can be positive, negative, or zero.

Problem 21

Find (∫_(-1)^( 1))3+(Sqrt[1-(x^2)])dx.

Solution

This is an integral which would be extremely difficult to do via Riemann sums (as we have no summation formula involving square roots) and difficult for us to do via the Fundamental Theorem (as it takes more work to find an antiderivative - it essentially requires a “reverse substitution” such as x=sin(θ)).  However if do the graph of y=3+(Sqrt[1-(x^2)]) it looks like this:

Plot[ 3+Sqrt[1-x^2], {x,-1,1}, AspectRatio->Automatic, Filling-> Axis, PlotLabel-> y==3+Sqrt[1-x^2], PlotRange->{0,4}]

From the picture we can see this is a semi-circle of radius 1 sitting on top of a rectange of base 2 and height 3.  Therefore the area is A=(1/2)π (r^2)+b h=(1/2)π (1^2)+2×3=(π/2)+6.

Note that you could also do this as (∫_(-1)^( 1))3+(Sqrt[1-(x^2)])dx=(∫_( -1)^( 1))3dx+(∫_( -1)^1)(Sqrt[1-(x^2)])dx, which might make it easier to recognize the rectangle and semi-circle separately as they are in distinct pieces.

Problem 22

Find (∫_( 0)^5)|x-2|dx.

Solution

Graphical solution:

The graph of y=|x-2| is a “V” shape with the corner of the “V” at the point (2,0):

Plot[ Abs[x-2], {x,0,5}, PlotLabel->TraditionalForm[ y==Abs[x-2]], Filling-> Axis]

So the area under the curve is that of 2 right triangles, one of base and height 2 and one of base and height 3.  So the area is given by A=(1/2)(Subscript[b, 1])(Subscript[h, 1])+(1/2)(Subscript[b, 2])(Subscript[h, 2])=(1/2)×2×2+(1/2)×3×3=2+(9/2)=(13/2).

Algebraic solution:

As |x-2|=({{{, ({{x-2, x≥2}, {2-x, x<2}})}})  we can split the integral up as:
 (∫_(  0)^5)|x-2|dx=(∫_( 0)^2)|x-2|dx+(∫_( 2)^5)|x-2|dx=(∫_( 0)^2)2-xdx+(∫_( 2)^5)x-2dx=(2x-((x^2)/2) |({{x=2}, {x=0}}))+(((x^2)/2)-2x |({{x=5}, {x=2}}))=((4-2)-(0-0))+(((25/2)-10)-(2-4))=(2-0)+((5/2)+2)=(13/2)

Problem 23

State both parts of the Fundamental Theorem of Calculus.

Solution

The first part states that if f(x) is continuous on [a,b], then (d/(dx))(∫_( t=a)^(t=x))f(t)dt=f(x) on (a,b).   That is, for a continuous function f(x) its area functions are antiderivatives (as a consequence, all continuous functions have antiderivatives).

The second part states that if F(x) in an antiderivative for f(x) on [a,b], then (∫_( x=a)^(x=b))f(x)dx=F(x)|({{x=b}, {x=a}})=F(b)-F(a).  This is what we use to find definite integrals if we can find an antiderivative.

Problem 24

Find (∫_( -1)^2)5+4x-7(x^5)dx.

Solution

(∫_( -1)^2)5+4x-7(x^5)dx=(5x+2(x^2)-(7/6)(x^6))|({{x=2}, {x=-1}})=(5(2)+2(((2))^2)-(7/6)(((2))^6))-(5(-1)+2(((-1))^2)-(7/6)(((-1))^6))=(10+12-(224/3))-(-5+2-(7/6))=-(105/2)
Note that this indicates in the graph of y=2sin(x)-x on [0,π] there is more area below the axis than above it:

Plot[ 5+4x-7x^5, {x,-1,2}, PlotRange->All, Filling-> Axis, FillingStyle-> {LightRed, Orange}, PlotLabel-> y==5+4x-7x^5
]

Problem 25

Find (∫_( 0)^π)2sin(x)-xdx.

Solution

(∫_( 0)^π)2sin(x)-xdx=(-2cos(x)-((x^2)/2))|({{x=π}, {x=0}})=(-2cos(π)-((π^2)/2))-(-2cos(0)-((0^2)/2))=(2-((π^2)/2))-(-2-0)=4-((π^2)/2)≈-0.934802.  Note that this indicates in the graph of y=2sin(x)-x on [0,π] there is more area below the axis than above it:

Plot[ 2Sin[x]-x, {x,0,Pi}, Filling->Axis, PlotLabel-> y==2Sin[x]-x, FillingStyle-> {LightRed, Orange}]

Problem 26

Find (∫_(  0)^( π/4))(sec^2)(x)+4sec(x)tan(x)dx.

Solution

(∫_(  0)^( π/4))(sec^2)(x)+4sec(x)tan(x)dx=(tan(x)+4sec(x))|({{x=π/4}, {x=0}})=(tan((π/4))+4sec((π/4)))-(tan(0)+4sec(0))=(1+4(Sqrt[2]))-(0+4)=4(Sqrt[2])-3.

Problem 27

State the Mean Value Theorem for Integrals

Solution

The Mean Value Theorem states that if f(x) is continuous on [a,b] then for at least one point x=c in [a,b] f(c)=(1/(b-a))(∫_( a)^b)f(x)dx - that is the graph must hit its average value at least once on a given interval.  It does not say exactly how many times it happens or how to find where it happens, just that it happens at least once.

Problem 28

Find the average value of (1/(x^3)) on [1,3].  Where does the function take on its average value?

Solution

The average value of (1/(x^3)) on [1,3] is (1/(3-1))(∫_(  1)^( 3))(1/(x^3))dx=(1/2)(∫_( 1)^3)(x^(-3))dx=(1/2)((x^(-2))/(-2))|({{3}, {1}})=-(1/(4(x^2)))|({{3}, {1}})=(-(1/36))-(-(1/4))=(1/4)-(1/36)=(2/9).  Setting (1/(x^3)) equal to (2/9) gives (1/(x^3))=(2/9) => 2(x^3)=9=>(x^3)=(9/2)=>x=((9/2)^(1/3))≈1.65096.  So this is where the function takes on its average value:

Show[Plot[ 2/9, {x,1,3}], Plot[1/x^3, {x,1,3}, Filling-> Axis], Graphics[ {PointSize[Medium],Point[{(9/2)^(1/3),2/9}]} ], PlotRange-> All, PlotLabel->Row[{ y==1/x^3, " and ", y==2/9}]]

Problem 29

Find the average value of y=1+2x-(x^2) on [-4,4].  Where does the function take on its average value?

Solution

The average value of y=1+2x-(x^2) on [-4,4] is given by the formula 
(1/(4-(-4)))(∫_(  -4)^( 4))1+2x-(x^2)dx=(1/8)(∫_( -4)^( 4))1+2x-(x^2)dx=(1/8)(x+(x^2)-((x^3)/3))|({{4}, {-4}})=(1/8)( (4+16-(64/3))-(-4+16-((-64)/3)))=(1/8)(8-(128/3))=1-(16/3)=-(13/3).  Setting 1+2x-(x^2)=-(13/3) is the same as (x^2)-2x-(16/3)=0=>x=((2±(Sqrt[4+(64/3)]))/2)=1±(Sqrt[(19/3)])≈-1.52, 3.52.  So this function hits its average value twice on [-4,4]:

Show[ Plot[ 1+2x-x^2, {x,-4,4}, Filling-> Axis, FillingStyle-> {LightRed, Orange}],Plot[-13/3, {x,-4,4}], Graphics[ {PointSize[Medium], Point[ {{1+Sqrt[19/3],-13/3},{1-Sqrt[19/3],-13/3}}]}],PlotLabel-> Row[ {TraditionalForm[ y==1+2x-x^2], " and ", TraditionalForm[y==-13/3]}] ]

Problem 30

Find (d/(dx))(∫_( 1)^( x))sin(3t+1)dt.  Find (d/(dx))(∫_( 2)^( x))sin(3t+1)dt.  Is this a coincidence?

Solution

First as sin(3x+1) is continuous everywhere by the Fundamental Theorem of Calculus we have  (d/(dx))(∫_( 1)^( x))sin(3t+1)dt=sin(3t+1)|({{ }, {t=x}})=sin(3x+1).

Then we also have (d/(dx))(∫_( 2)^( x))sin(3t+1)dt=sin(3t+1)|({{ }, {t=x}})=sin(3x+1).

This makes perfect sense as (∫_( 1)^( x))sin(3t+1)dt-(∫_( 2)^( x))sin(3t+1)dt=(∫_( 1)^2)sin(3t+1)dt, which is a number.  As the two area functions differ by a constant they should have the same derivative.

Problem 31

Find (d/(dx))(∫_( x)^4)(t^2)-2t+(1/t)dt (assume x>0)

Solution

By the properties of definite integrals, (∫_( x)^4)(t^2)-2t+(1/t)dt=-(∫_( 4)^x)(t^2)-2t+(1/t)dt.  So (d/(dx))(∫_( x)^4)(t^2)-2t+(1/t)dt=(d/(dx))-(∫_(  4)^( x))  (t^2)-2t+(1/t)dt=-(d/(dx))(∫_( 4)^x) (t^2)-2t+(1/t)dt.  As (t^2)-2t+(1/t) is continuous on the positive real numbers this becomes -(d/(dx))(∫_( 4)^x) (t^2)-2t+(1/t)dt=-((t^2)-2t+(1/t)|({{ }, {t=x}}))=-((x^2)-2x+(1/x))=-(x^2)=2x-(1/x).

Problem 32

Find (d/(dx))(∫_( 1)^( sin(x)))(1/((t^2)+1))dt and (d/(dx))(∫_(x^3)^(sin(x)))(1/((t^2)+1))dt.

Solution

As (1/((t^2)+1)) is continuous everywhere by the chain-rule-enchanced version of the Fundamental Theorem of Calculus we have (d/(dx))(∫_( 1)^( sin(x)))(1/((t^2)+1))dt=((1/((t^2)+1))|({{ }, {t=sin(x)}}))(d/(dx))sin(x)=(1/((sin^2)(x)+1))cos(x).

By using the version that has functions in both limits ((d/(dx))(∫_(g(x))^(h(x)))f(t)dt=f(h(t))h'(t)-f(g(x))g'(x) ), we have  (d/(dx))(∫_(x^3)^(sin(x)))(1/((t^2)+1))dt=((1/((t^2)+1))|({{ }, {t=sin(x)}}))(d/(dx))sin(x)-((1/((t^2)+1))|({{ }, {t=(x^3)}}))(d/(dx))(x^3)=(1/((sin^2)(x)+1))cos(x)-(1/(((((x^3)))^2)+1))3(x^2)=((cos(x))/((sin^2)(x)+1))-((3(x^2))/((x^6)+1)).

Problem 33

Find ∫((x^2)/((((x^3)+5))^3))dx.

Solution

This one does not have the form of any of our basic integral rules which suggests we should try integration by substitution.  The complicated part is the (x^3)+5 in the denominator, so try u=(x^3)+5.  Then du=3(x^2)dx, or (x^2)dx=(1/3)du.  Putting this into the integral gives us ∫((1/3)/(u^3))du=(1/3)∫(u^(-3))du=(1/3)((u^(-2))/(-2))+C=-(1/(6(u^2)))+C=-(1/(6((x^3)+5)))+C.

Problem 34

Find ∫(x^4)+cos(x)sin(sin(x))dx.

Solution

We can handle the first term of the integral easily - splitting it off gives us ∫(x^4)+cos(x)sin(sin(x))dx=∫(x^4)dx+∫cos(x)sin(sin(x))dx=((x^5)/5)+∫cos(x)sin(sin(x))dx.   In the remaining integral (which has the “+C ” embedded in it still) we can use integration by substitution - use u=sin(x), du=cos(x)dx.  This will give us ((x^5)/5)+∫sin(u)du=((x^5)/5)-cos(u)+C=((x^5)/5)-cos(sin(x))+C.

Problem 35

Find ∫(sec^2)(3t+1)dt

Solution

This is almost the rule that yields tangent, so use integration by substitution with u=3t+1, d u=3 d t (or dt=(1/3)du).  This will give us ∫(sec^2)(3t+1)dt=∫(1/3)(sec^2)(u)du=(1/3)tan(u)+C=(1/3)tan(3t+1)+C.

Problem 36

Find (∫_(  0)^2)(((x^3)(4+5(x^4)))^7)dx.

Solution

While in theory we could multiply this out and do the integral directly, it is easier here to use definite integration by substitution.  Letting u=4+5(x^4), we would have du=20(x^3)dx (or (x^3)dx=(1/20)du), x=0=>u=4, and x=2 =>u=84.  So (∫_(  x=0)^(x=2))(((x^3)(4+5(x^4)))^7)dx=(1/20)(∫_( u=4)^(u=84))(u^7)du=(1/20)((u^8)/8)|({{u=84}, {u=4}})=((84^8)/160)-((4^8)/160)=15492243193856.  

Note that while this is a lot easier than multiplying out (((x^3)(4+5(x^4)))^7)=16384 (x^3)+143360 (x^7)+537600 (x^11)+1120000 (x^15)+1400000 (x^19)+1050000 (x^23)+437500 (x^27)+78125 (x^31) by hand.

Problem 37

Find (∫_(  0)^( π/4))((sin(x))/((cos^4)(x)))dx.

Solution

The integrand is complicated enough to try integration by substitution with u=cos(x). Then du=-sin(x)dx (or sin(x)dx=-du), x=0 => u=cos(0)=1, and x=(π/4) => u=cos((π/4))=(1/(Sqrt[2])).  So 
(∫_(  x=0)^( x=π/4))((sin(x))/((cos^4)(x)))dx=-(∫_( u=1)^( u=1/(Sqrt[2])))(1/(u^4))du=(Subscript[∫ , u=1/(Sqrt[2])]^( u=1))(1/(u^4))du=(Subscript[∫ , u=1/(Sqrt[2])]^( u=1)) (u^(-4))du=-(1/3)(u^(-3))|({{u=1}, {u=(1/(Sqrt[2]))}})=-(1/(3(u^3)))|({{u=1}, {u=(1/(Sqrt[2]))}})=(-(1/3))-(-(1/(3/(2(Sqrt[2])))))=((2(Sqrt[2]))/3)-(1/3)=((2(Sqrt[2])-1)/3)

Problem 38

Find (∫_(  0)^1)(1/(((5-x))^2))dx

Solution

Try definite integration with u=5-x (note that as x goes from 0 to 1 we have 5-x≠0 so the integrand is 
continuous).  Then du=-dx (or dx=-du), x=0 => u=5, and x=1 => u=4.  So (∫_(  x=0)^(x=1))(1/(((5-x))^2))dx=-(∫_( u=5)^(u=4))(1/(u^2))du=-(∫_( u=5)^(u=4)) (u^(-2))du=-((u^(-1))/(-1))|({{u=4}, {u=5}})=(1/u)|({{u=4}, {u=5}})=(1/4)-(1/5)=(1/20).

Problem 39

Find (∫_(  0)^(1/2))((x^3)/(Sqrt[1-(x^4)]))dx.

Solution

Try integration by substitution with u=1-(x^4) (note that as x goes from 0 to (1/2), 1-(x^4)≠0 so the integrand is continuous).  Then du=-4(x^3)dx (or (x^3)dx=-(1/4)du),  x=0 => u=1, and x=(1/2) => u=1-((((1/2)))^4)=(15/16).  Then (∫_(  x=0)^(x=1/2))((x^3)/(Sqrt[1-(x^4)]))dx=-(1/4)(∫_( u=1)^(u=15/16)) (1/(Sqrt[u]))du=(1/4)(∫_( u=15/16)^(u=1)) (u^(-1/2))du=(1/4)(((u^(1/2))/(1/2)))|({{u=1}, {u=(15/16)}})=(1/2)(Sqrt[u])|({{u=1}, {u=(15/16)}})=(1/2)(Sqrt[1])-(1/2)(Sqrt[(15/16)])=(1/2)-(1/2)((Sqrt[15])/4)=(1/2)-((Sqrt[15])/8)=((4-(Sqrt[15]))/8)

Problem 40

Find (∫_(-2)^2)((2x)/((x^2)+1))dx.

Solution

Try the substitution u=(x^2)+1.  Then du=2xdx (or xdx=(1/2)du), x=-2 => u=5, and x=2 => u=5.  So the integral becomes  (∫_( x=-2)^(x=2))((2x)/((x^2)+1))dx=(1/2)(∫_( u=5)^(u=5))(1/u)du.  At first glance it looks like we are stuck, as we don’t have an antiderivative formula for (1/u) ((1/u)=(u^(-1)), which is the hole in the power rule).  However as both the upper and lower limit of integration are 5 the integral must be 0, so (∫_( x=-2)^(x=2))((2x)/((x^2)+1))dx=(1/2)0=0!

If you graph f(x)=((2x)/((x^2)+1)) you can see what is happening in terms of symmetry.  f(-x)=((2(-x))/((((-x))^2)+1))=-((2x)/((x^2)+1))=-f(x), so f(x) has the “odd” symmetry - that is, the graph is symmetric about the origin:

Plot[ 2x/(x^2+1), {x,-2,2}, Filling->Axis, FillingStyle-> {LightRed, Orange}, PlotLabel-> y==2x/(x^2+1)]

As we are integrating on an interval centered at x=0 the symmetry in the graph forces the positive and negative areas to cancel out exactly, leaving a net area of 0.  This wouldn’t work on the interval [1,3], but it would work on [-3,3] or in general any interval [-a,a] (provided the function with the odd symmetry is continuous).