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!).
Please read these answers after you work on the problems. Only 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 1
A <--> 6: at 6, f is increasing (f´(6) > 0) and concave up
(f´´(6) >0).
B <--> 3: 3 is a critical number (f´(3) = 0) and f is concave down
(f´´(3)<0).
C <--> 2: at 2 f is decreasing (f´(2)< 0) and concave up
(f´´(2)>0).
D <--> 4: at 4 f is increasing (f´(4)>0) and concave down
(f´´(4)>0).
E <--> 0: at 0 f is decreasing (f´(0)<0) and concave down
(f´´(0)<0).
F <--> 1: f has a critical number at 1 (f´(1)=0) and f is concave up
(f´´(1)>0).
G <--> 7: f has a critical number at 7 (f´(7)>0) and f has an
inflection point at 7 (f´´(7)=0).
H <--> 5: at 5 f is decreasing (f´(5)<0) and f has an inflection
point at 5 (f´´(5)=0).
Note that an x with f´´(x)=0 may not be an
inflection point (consider f(x)=x4 at 0 for example). But
here we're told that the PIctures and the x's match up, so the
correspondances 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!).
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*32 + A*3*2*2*(-1) + B* 3* 22*(-1)=0,
or -3A-12B=-6.
So we need to solve the pair of equations 12A+8B=-8 and -3A-12B=-6. We
can make them friendlier by dividing the first one by 4 and the second
by -3. The pair of equations becomes 3A+2B=-2 and A+4B=2. Double the
first and subtract the second to get 5A=-6 so A=-6/5 and then B= 4/5.
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 part has antiderivative -5cos(x). The other part 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
antidervative 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). 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)22 +2C+D=0. So D=0 from the first equation, and the second equation becomes -18+2D=0 so D=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 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. So we need to compute and then "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*42)/(33))=ln(32/27).
Remark 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)=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 seems a bit more involved than the preceding
problem. So here we go: using the substitution u=1+3x3
gives 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 then
F(ee)-F(e)=ln(ln(ee))-ln(ln(e))= ln(e
ln(e))-ln(1)=ln(e)-0=1.
Problem 10 a) If G(x)=exsin(x)-excos(x),
then
G´(x)=exsin(x)+excos(x)-excos(x)+exsin(x)
with two uses of the product rule. The derivative simplifies to
2exsin(x).
b) If we want to use the FTC, we need a function whose derivative is
exsin(x). Fortunately (!) reading the result of a) provides
an example: merely divide the function g by 2. So an antiderivative of
exsin(x) is
F(x)=(1/2)(exsin(x)-excos(x)), and the FTC tells
us that the value of the definite integral is just F(PI)-F(0). Here
F(PI)=(1/2)(ePIsin(PI)-ePIcos(PI))=(1/2)ePI
and F(0)=(1/2)(e0sin(0)-e0cos(0))=-(1/2). So the
definite integral is (1/2)ePI+(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(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.
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. The truly compulsive would answer (63)/6 (or maybe (31)/2).
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(x2-2x+1= x3-2x2+x. An antiderivative of this is (1/4)x4-(2/3)x3+(1/2)x2=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.