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 e^{Frog}. 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)=x^{4} at 0 for example). But
here we're told that the pictures and the x's match up exactly, so the
correspondances given above are the only possible answers.

**Problem 2** y=(2x+1)exp(-x^{2}). y=0 only when x=-(1/2).
y´=2exp(-x^{2})+(2x+1)exp(-x^{2})(-2x)=(-4x^{2}-2x+2)exp(-x^{2}). y´=0
only when x=-1 or
x=1/2. y´´=(-8x-2)exp(-x^{2})+(-4x^{2}-2x+2)exp(-x^{2})(-2x)=(8x^{3}+4x^{2}-12x-2)exp(-x^{2}). When
is this equal to 0? The exponential part is never 0, so we need to
know when 8x^{3}+4x^{2}-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!).

- There is a maximum at x=1/2 (this is a relative and absolute maximum).
- There is a local minimum at x=-1 (this is a relative and absolute minimum).
- The function is decreasing in the intervals (-infinity,-1] and [1/2,infinity).
- The function is increasing in the interval [-1,1/2].
- The function is concave up in the intervals [-1.4,-.2] and [1.1,infinity).
- The function is concave down in the intervals (-infinity,-1.4] and [-.2,1.1].
- There are points of inflection at x=-1.4 and x=-.2 and x=1.1 (approximately).

**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 3^{2}
+ A*3*2^{2} + B 2^{3} =1, or 12A+8B=-8.

We can differentiate the equation implicitly to get: 2x+Ay^{2}
+ Ax*2y y´ + B*3y^{2}y´=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*2^{2} + A*3*2*2*(-1) + Bo* 3* 2^{2}*(-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)x^{3}-3e^{x}+5ln(x)+C.

b) First expand the square. The integrand (the function to be
integrated) is then
(x^{3})^{2}+2x^{3}5+5^{2}=x^{6}+10x^{3}+25. This
can be integrated again piece by piece, and the answer is
(1/7)x^{7}+(10/4)x^{4}+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=x^{2}+5 so du=(2x)dx, and
(1/2)du=xdx. Then the integrand becomes (1/2)(1/u)du, so an
antidervative is (1/2)ln(u) +C which is (1/2)ln(x^{2}+5)+C.

**Problem 6** All antiderivatives of 2x^{3}-1 are of the
form (1/2)x^{4}-x+C. To find C we use the initial condition,
inserting x=2 and y=3 in the equation y=(1/2)x^{4}-x+C. This
gives us 3=(1/2)2^{4}-2+C, so 3=6+C or C=-3. Therefore the
answer is y=(1/2)x^{4}-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)x^{2}+Cx +D (there will be two possibly different
constants, C and D, one for each antidifferentiation). Then 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)2^{2}
+2C+D=0. So D=0 from the first equation, and the second equation
becomes -18+2C=0 so C=9. The function K(x) is
-(9/2)x^{2}+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.

** Apology & misprint alert ** Please note that the statement of
the problem is *supposed* to include "K(0)=0 and K(2)=0" * not
* K´(2)=0. I've corrected this in the web statements of the
problem, but obviously cannot reach out and delete the ´ from
every printed version! I regret any confusion which may occur. Thanks
to student KPL for pointing out this error! Of course, the problem is
still interesting in the version with the ´ . Then -9(2)+C=0 so
C=18, and K(x)= -(9/2)x^{2}+18x+D, and K(0)=0 makes D=0 again,
so K(x)= -(9/2)x^{2}+18x. This has a maximum when x=2, and
here K(2)=18. In fact, maybe this version of the problem is * more
* interesting!

**Problem 8** The integral from 1 to 4 is equal to the sum of the
integrals from 1 to 2 and from 2 to 4 (just draw a picture of areas to
convince yourself of this!). So the integral from 1 to 2 is the
difference: the integral from 1 to 4 minus the integral from 2 to
4. We need to "simplify" a bunch of logs>

(ln(2)-2ln(3)-ln(5)+2ln(6))-(ln(3)-2ln(4)-ln(5)+2ln(6))=
ln(2)-3ln(3)+2ln(4)=(using properties of
logs)ln((2*4^{2})/(3^{3}))=ln(32/27).

**Comment** It is *not* completely obvious (!) where the
formulas in this problem came from. In fact, they came from a use of
the FTC, because the derivative of -ln(x+1)+2ln(x+2) turns out to be
(after some algebra!) x/((x+1)(x+2)). Of course getting such a formula
is, in turn, not obvious, but it and others like it (such as the one
implied in the next problem) can be explained.

**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)=x^{3}-x^{-4} so
F(x)=(1/4)x^{4}-(1/-3)x^{-3}, and the value of the
integral is
F(2)-F(1)=((1/4)2^{4}+(1/3)2^{-3})-((1/4)1^{4}+(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 4e^{2x}. We can guess
(2e^{2x}!) or, better, be guided by a substitution: u=2x,
du=2dx, so (1/2)du=dx and 4e^{2x}dx=2e^{u}du. This
antidifferentiates to 2e^{u}=2e^{2x}. We'll use the
last answer as our F(x). The definite integral is then
F(ln(3))-F(0)=2e^{2(ln(3))}-2e^{2(0)}=2e^{ln(9)}-2(1)=2(9)-2=16.

c) Here to apply the FTC we need an antiderivative of
x^{2}(1+3x^{3})^{(1/2)}. This can also be done
with a substitution, but seems a bit more involved than the preceding
problem. So here we go, using the substitution u=1+3x^{3}
gives du=9x^{2}dx so when we divide by 9 we get
(1/9)du=x^{2}dx, and
x^{2}(1+3x^{3})^{(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)u^{3/2}. Back now to x's:
(1/9)(2/3)u^{3/2}=(1/9)(2/3)(1+3x^{3})^{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)2^{3})^{3/2}-(1/9)(2/3)(1+(3)0^{3})^{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=e^{x} so
du=e^{x}dx and e^{x}sin(e^{x})dx=sin(u)du,
which has antiderivative -cos(u)=-cos(e^{x})=F(x). The answer
we want is F(ln(2PI))-F(ln(PI))=-cos(e^{ln(2PI)})-(-cos(e^{ln(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 then
F(e^{e})-F(e)=ln(ln(e^{e}))-ln(ln(e))= ln(e
ln(e))-ln(1)=ln(e)-0=1.

**Problem 10** a) If g(x)=e^{x}sin(x)-e^{x}cos(x),
then
g´(x)=e^{x}sin(x)+e^{x}cos(x)-e^{x}cos(x)+e^{x}sin(x)
with two uses of the product rule. The derivative simplifies to
2e^{x}sin(x).

b) If we want to use the FTC, we need a function whose derivative is
e^{x}sin(x). Fortunately (!) reading the result of a) provides
an example: merely divide the function g by 2. So an antiderivative of
e^{x}sin(x) is
F(x)=(1/2)(e^{x}sin(x)-e^{x}cos(x)), and the FTC tells
us that the value of the definite integral is just F(PI)-F(0). Here
F(PI)=(1/2)(e^{PI}sin(PI)-e^{PI}cos(PI))=(1/2)e^{PI}
and F(0)=(1/2)(e^{0}sin(0)-e^{0}cos(0))=-(1/2). So the
definite integral is (1/2)e^{PI}+(1/2).

**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(x^{2}))/(1+x^{4}). 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
x^{5} from 1 to 2. We evaluate the integral using the FTC. An
antiderivative of x^{5} is (1/6)x^{6}=F(x). The
integral's value is F(2)-F(1), which is (1/6)2^{6}
-(1/6)1^{6}, a fine answer. This is equal to (63)/6 or even
(31)/2 or 15.5 or ...

**Problem 14** Graphing y=x(x-1)^{2} (or just thinking
about it a bit) reveals that the "arch" is for x's between 0 and 1. So
we must compute the definite integral of x(x-1)^{2} from 0 to
1. To apply the FTC it is useful to rewrite x(x-1)^{2}: it is
x(x^{2}-2x+1)= x^{3}-2x^{2}+x. An
antiderivative of this is
(1/4)x^{4}-(2/3)x^{3}+(1/2)x^{2}=F(x). Finally,
the integral's value is F(1)-F(0)= (1/4)-(2/3)+(1/2), a fine
answer. If you want to be more confident that the answer is positive,
a bit of arithmetic reveals that it is 1/12.