The answers ...

Warning! Please read these answers after you work on the problems. Reading the answers without working on the problems is not a useful strategy!

Notation exp(Frog) will denote eFrog. sqrt(Toad) will denote the square root of Toad (so sqrt(4) is 2). * will be used for multiplication, so 3*4 is 12. "FTC" is an abbreviation for "Fundamental Theorem of Calculus".

Problem 0
C=9/4 and D=ln(4/3).

Problem 1
A <--> 6: at 6, Q is increasing (Q´(6) > 0) and concave up (Q´´(6) >0).
B <--> 3: 3 is a critical number (Q´(3) = 0) and Q is concave down (Q´´(3)<0).
C <--> 2: at 2 Q is decreasing (Q´(2)< 0) and concave up (Q´´(2)>0).
D <--> 4: at 4 Q is increasing (Q´(4)>0) and concave down (Q´´(4)>0).
E <--> 0: at 0 Q is decreasing (Q´(0)<0) and concave down (Q´´(0)<0).
F <--> 1: Q has a critical number at 1 (Q´(1)=0) and Q is concave up (Q´´(1)>0).
G <--> 7: Q has a critical number at 7 (Q´(7)=0) and f has an inflection point at 7 (Q´´(7)=0).
H <--> 5: at 5 Q is decreasing (f´(5)<0) and Q has an inflection point at 5 (Q´´(5)=0).
Note that an x with Q´´(x)=0 may not be an inflection point (consider Q(x)=x4 at 0 for example). But here we're told that the pictures and the x's match up exactly, so the correspondences given above are the only possible answers.

Problem 2 y=(2x+1)exp(-x2). y=0 only when x=-(1/2). y´=2exp(-x2)+(2x+1)exp(-x2)(-2x)=(-4x2-2x+2)exp(-x2). y´=0 only when x=-1 or x=1/2. y´´=(-8x-2)exp(-x2)+(-4x2-2x+2)exp(-x2)(-2x)=(8x3+4x2-12x-2)exp(-x2). When is this equal to 0? The exponential part is never 0, so we need to know when 8x3+4x2-12x-2=0. Although there is a formula for solving cubic equations, the roots can be approximated by graphing and looking: y´´=0 for x=-1.4 and x=-.2 and x=1.1 (approximately!).
All the information desired can be gotten from these computations (and from looking at the signs of the derivatives on each side of where they are 0!).

It is also true that the x-axis (y=0) is a horizontal asymptote.

Problem 3 The line goes through (4,f(4)) so f(4)=3*4+7=19. The slope of the line is the derivative of f at 4, so f´(4)=3.

Problem 4 Since the curve goes through (3,2), the pair of values x=3 and y=2 must satisfy the equation. Therefore 32 + A*3*22 + B 23 =1, or 12A+8B=-8.
We can differentiate the equation implicitly to get: 2x+Ay2 + Ax*2y y´ + B*3y2y´=0. When x=3 and y=2, y´ (which is the slope of the tangent line) is -1. Therefore we also know that 2(3)+A*22 + A*3*2*2*(-1) + Bo* 3* 22*(-1)=0, or -8A-12B=-6.
So we need to solve the pair of equations 12A+8B=-8 and -8A-12B=-6. We can make them friendlier by dividing the first one by 4 and the second by -2. The pair of equations becomes 3A+2B=-2 and 4A+6B=3. Triple the first and subtract the second to get 5A=-9 so A=-9/5=-1.8 and then B=1.7.

Problem 5 a) Each piece is easy. The answer is (7/3)x3-3ex+5ln(x)+C.
b) First expand the square. The integrand (the function to be integrated) is then (x3)2+2x35+52=x6+10x3+25. This can be integrated again piece by piece, and the answer is (1/7)x7+(10/4)x4+25x+C.
c) The first piece has antiderivative -5cos(x). The other piece will need a simple substitution with u=5x, so du = 5dx and (1/5)du=dx etc. so the answer will be (1/5)sin(u)=(1/5)sin(5x). The combined answer is -5cos(x)+(1/5)sin(5x)+C.
d) Here use the substitution u=x2+5 so du=(2x)dx, and (1/2)du=xdx. Then the integrand becomes (1/2)(1/u)du, so an antiderivative is (1/2)ln(u) +C which is (1/2)ln(x2+5)+C.

Problem 6 All antiderivatives of 2x3-1 are of the form (1/2)x4-x+C. To find C we use the initial condition, inserting x=2 and y=3 in the equation y=(1/2)x4-x+C. This gives us 3=(1/2)24-2+C, so 3=6+C or C=-3. Therefore the answer is y=(1/2)x4-x-3.

Problem 7 If K´´(x)=-9, then K´(x)=-9x+C by antidifferentiating once. Doing it again we get K(x)=-(9/2)x2+Cx +D (there will be two possibly different constants, C and D, one for each antidifferentiation). We can get one equation involving C and D from each of the two given equations. So K(0)=0 gives D=0 and K(2)=0 gives -(9/2)22 +2C+D=0. D=0 from the first equation, and the second equation becomes -18+2C=0 so C=9. The function K(x) is -(9/2)x2+9x. The derivative of K is 0 at 1, and there K has its maximum value, which is 9/2. This is also the highest the curve can be.

Problem 8 We are told that the integral of 4/(1+x2) from 0 to 1 is PI. If we multiply by 1/4, we learn that the integral of 1/(1+x2) from 0 to 1 is PI/4.
We are told that the integral of 3/(1+x2) from 0 to sqrt(3) is PI. If we multiply by 1/3, we learn the the integral of 1/(1+x2) from 0 to sqrt(3) is PI/3.
The integral from 0 to sqrt(3) is the sum of the integral from 0 to 1 and the integral from 1 to sqrt(3). Therefore the integral of 1/(1+x2) from 1 to sqrt(3) must be (PI/3)-(PI/4), which is PI/12.

Comment It is not clear where the formulas in this problem came from. In fact, they are "easy" if you learn a bit about the calculus of the inverse trigonometric functions, which was not covered in this course (don't panic!). It turns out that the derivative of arctan x (the inverse function to tangent) is 1/(1+x2). That this derivative is a rather simple algebraic function is not at all predictable from a naive view of the trig functions. Knowing this (again, not part of Math 135!) the facts which are given in this problem are not difficult to verify.

Problem 9 We'll use the FTC on all of these. The general outline is: to compute the definite integral of f from a to b, find an antiderivative F of f (that is, a function F with F´ =f), and then the value of the definite integral is F(b)-F(a). a) Here f(x)=x3-x-4 so F(x)=(1/4)x4-(1/-3)x-3, and the value of the integral is F(2)-F(1)=((1/4)24+(1/3)2-3)-((1/4)14+(1/3)1-3) which is a fine answer. If you want to do some unnecessary arithmetic, you'll get (I hope!) (83)/(24) as "the answer".
b) We need an antiderivative of 4e2x. We can guess (2e2x!) or, better, be guided by a substitution: u=2x, du=2dx, so (1/2)du=dx and 4e2xdx=2eudu. This antidifferentiates to 2eu=2e2x. We'll use the last answer as our F(x). The definite integral is then F(ln(3))-F(0)=2e2(ln(3))-2e2(0)=2eln(9)-2(1)=2(9)-2=16.
c) Here to apply the FTC we need an antiderivative of x2(1+3x3)(1/2). This can also be done with a substitution, but it is a bit more involved than the preceding problem. If we substitute u=1+3x3, then du=9x2dx so when we divide by 9 we get (1/9)du=x2dx, and x2(1+3x3)(1/2)dx becomes (in terms of u) just (1/9)u(1/2)du. The last is just a constant multiplying a power, so we can write an antiderivative for it: (1/9)(2/3)u3/2. Back now to x's: (1/9)(2/3)u3/2=(1/9)(2/3)(1+3x3)3/2 and use this last answer as F(x). By the FTC, the integral equals F(2)-F(0)=(1/9)(2/3)(1+(3)23)3/2-(1/9)(2/3)(1+(3)03)3/2. This is a fine answer. If you would like to clean it up ("simplify"), you should get (248)/(27).
d) Here the substitution is easy: try u=ex so du=exdx and exsin(ex)dx=sin(u)du, which has antiderivative -cos(u)=-cos(ex)=F(x). The answer we want is F(ln(2PI))-F(ln(PI))=-cos(eln(2PI))-(-cos(eln(PI)))=-cos(2PI)+cos(PI)=-1-1=-2.
e) Finding an antiderivative here can be a little tricky, because it is almost too easy! We need to cope with 1/(x ln(x)). Try u=ln(x). Then du=(1/x)dx, and (1/(x ln(x)))dx=(1/u)du. An antiderivative of this is ln(u), so going back to x's we get F(x)=ln(ln(x)). If you don't believe the result, check it by differentiating! The value of the integral is therefore F(ee)-F(e)=ln(ln(ee))-ln(ln(e))= ln(e ln(e))-ln(1)=ln(e)-0=1.

Problem 10 a) The derivative of Ex2 ex +Fxex +Gex is supposed to be x2ex (differentiation will cause the "+ any constant" to vanish). Remember that E and F and G are supposed to be constants (specific numbers). Then the needed derivative is produced using the usual differentiation rules, including the product rule. The derivative is:


Wow! We can rewrite this algebraically, stressing the separate functions x2ex and xex and ex. The formula then becomes:

Ex2ex + (2E+F)xex + (F+G)ex

How could we get x2ex out of this? E should be 1. But then 2E+F should be 0, so that we'd better choose F=-2. And finally F+G should be 0, so that G=2.

b) One antiderivative of x2ex must be x2ex-2xex+2ex, using the result of a). The FTC then applies, and the value of the definite integral from 0 to 1 of x2ex is (x2ex-2xex+2ex evaluated at 1) - (x2ex-2xex+2ex evaluated at 0). This is e-2.

Problem 11 F(-42) is the integral from -42 to -42 of something: that's 0 because the upper and lower limits are the same. The other two questions use part of the FTC directly. If F(x) is the integral of f(t) from -42 to x, then the FTC says that F'(x)=f(x). Therefore F´(x) will be (sin(x2))/(1+x4). So F´(0) will be 0 since sin(0) is 0, and F´(sqrt(PI)) is also 0 since sin(PI)=0.

Problem 12 This problem tests yet again the FTC, and it also asks if you know the relationship of the definite integral to area. Remember that the definite integral of f on the interval [a,b] is the signed area "under" f in relation to the horizontal axis: count the area "above" the axis positively and the area "below" the axis negatively. The net result is what is reported by the definite integral. Also remember that g´ = f.
a) g(-1)=0: there's no area yet! g(0)=PI/4, the area of a quarter circle. g(5)=PI/2, because the positive area of a half circle from 1 to 3 cancels out the negative area of a half circle from 3 to 5, and we have the area of the half circle from -1 to 1 remaining. Part of the FTC allows us to compute g´ easily. g´(-1)=f(-1)=0 and g´(0)=f(0)=1 and g´(5)=f(5)=0.
b) g(3) is the definite integral of f from -1 to 3. The Riemann sum approximation to this with the specified partition and sample points is f(-1)(0-(-1))+f(0)(1-0)+f(1)(3-1)=0(1)+1(1)+0(2)=1.

Problem 13 The area is equal to the definite integral of x5 from 1 to 2. We evaluate the integral using the FTC. An antiderivative of x5 is (1/6)x6=F(x). The integral's value is F(2)-F(1), which is (1/6)26 -(1/6)16, a fine answer. This is equal to (63)/6 or even (31)/2 or 15.5 or ...

Problem 14 The area enclosed by sin(x) and the x-axis and the line y=PI/2 is just the definite integral of sin(x) from 0 to PI/2. An antiderivative of sin(x) is -cos(x), so that the FTC applies, and the area must be -cos(PI/2)-(-cos(0))=-0-(-1)=1.

Half the area is 1/2, of course. If the vertical line y=B divides this area into two equal parts, then the definite integral of sin(x) from 0 to B must equal 1/2. The FTC says that the definite integral of sin(x) from 0 to B is (-cos(x) evaluated at B)-(-cos(x) evaluated at 0). This is -cos(B)+1=(1/2), so that cos(B)=1/2, and since B is between 0 and PI/2, B must be PI/3.

Maintained by and last modified 4/22/99.