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".
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).
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
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
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
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),
with two uses of the product rule. The derivative simplifies to
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
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.