A1.   If f(x) = log2 (5x), ind f/ (x).

Solution:  We know that the derivative of logb x is 1xlnb , and perhaps you remember that the derivative of logb cx for any constant c is also 1xlnb , which gives f/ (x) = 1xln2 .  But if you didn’t remember that, you can easily get it, either by simplifying f(x) before diferentiating:

f(x) = log2 5x = log2 5 + log2 x ⇒ f/ (x) = 0 + 1xln2

or by using the chain rule:

f/ (x) = ddx [log2 5x] = ddx[5x]5xln2 = 55xln2 = 1xln2

A2.   If y = e3 , ind

dy

dx .

Solution:  Since e3  is a constant, dy = 0.

A3.   If f(x) = tan(2x), ind f/  ( T6).

Solution:  We use the chain rule, along with the fact that the derivative of tan x is sec2 x, to get

f/ (x) = ddx [tan2x] =(sec2 2x)  [ ddx (2x)] = (sec2 2x)(2) = 2 sec2 2x

And now we evaluate this at x = 6:

f/  (T6) = 2 sec2 2 (T6) = 2 (sec T3)2  = 2 ( 1cos3)2  = 2 ( 12(1/2)2) = 2(4) = 8

A4.   If f/ (x) = sinx and f(0) = 2, ind f(x).

Solution:

\

So f(0) = 2 gives − cos0+C = 2, so − 1+C = 2 and thus C = 3. Therefore f(x) = − cosx+3 = 3− cos x.

A5.   Find \ dx.

Solution:

dx = \ (3(3)x(x) + 3x(1)) dx = \ (1 + 3x(1)) dx = \ 1 dx +3(1) \x(1) dx = x +3(1)(ln |x|) + C

A: tan4 x sec3 x + C     B: 12 tan4 x sec3 x + C        C: 4 sec3 x + C

D: 3 tan4 x + C            E: None of A, B, C or D

Solution:  We use the substitution t = tanx, which gives dt = sec2 xdx, so we get

\ 12 tan3 xsec2 xdx = \ 12t3 dt = 14(2t)4  + C = 3t4 + C = 3 tan4 x + C

A7.   Evaluate dx.

Solution: We again need the substitution rule. Letting t = 1+ x3 gives dt = 3x2 dx so that x2 dx =‘13) dt. And we see that when x = 0, t = 1, while x = 1 gives t = 2. Therefore we get

\0 1 dx    =   \1 2 1t2  ( 13) dt = 13 \1 2 t−2 dt = 13 [ ]1(2)  = 13 [−1t]1(2) =   13 [( − 12)− (− 11)] = 13 ( 1 − 12) = 13 × 12 = 16

A8.   Evaluate \1 e (ln x)3x dx.

Solution:  Letting t = ln x gives dt = 1x dx, so we have

\1 e (lnx)3x dx = \ t3 dt = \0 1 t3 dt = [ t44]0(1)  = 144 − 044 = 14

A9.   What is the average value of f(x) = x2 + 1 on the interval [0, 3]?

Solution:  Since the average value of f(x) on the interval [a,b] is given by \ab f(x)dx, in this case

\0 3 (x2 + 1) dx = 13 [ x33 + x]0(3)  = 13 [333 + 3 − (033 + 0)] = 13 (9 + 3 − 0) = 123 = 4

A10.   Find the area of the region between y = 1x and y = 0, from x = 2 to x = 5.

Solution:  Since 1北 > 0 when x > 0, the curve y = 1北 lies above the x-axis (i.e. y = 0) between x = 2 and x = 5 so we get

area  = \2 5 1x dx = [ln|x|]2(5)  = ln5 − ln2 = ln ( 52)

A11.   Find the area of the region between y = ^x and the y-axis, from y = 1 to y = 2.

Solution:  y = ^x is the upper half of the parabola x = y2 , so the region has this curve as right boundary, with the y-axis (i.e.  x = 0) as the left boundary, and the lines y = 1 and y = 2 as lower and upper boundaries, respectively. Using horizontal slicing we see that

area  = \1 2 y2 dy = [ y33]1(2)  = 233 − 133 = 83 − 13 = 73

A12.   Find the volume of the solid obtained when the region bounded by y = x2 , the x-axis and x = 2 is revolved

about the y-axis.

Solution:  Since we are revolving the region about a vertical axis, we slice horizontally and integrate with respect to y .  The region extends from x = 0 to x = 2, so we are only interested in the right half of the parabola y = x2 , which can be expressed as x = ^y .  This is the left and upper boundary of the region, which has y = 0 as lower boundary and x = 2 as right boundary. The left and lower boundaries intersect at y = 0, while the right and upper boundaries intersect when 2 = ^y , i.e. when y = 4. When a horizontal slice of this region is revolved about the y-axis, it produces a washer with outer radius R = 2 and inner radius r = ^y , so when the whole region is revolved, the resulting solid has volume

π \0 4  (22 − (^y)2 ) dy = π \0 4 (4 − y)dy = π [4y − y22]4  = π [( 16 − 162)− (0 − 022)] = π(16 − 8) = 8π

A13.   Find the volume of the solid obtained when the region bounded by y = x2 , the x-axis and x = 2 is revolved

about the x-axis.

Solution:  We have the same region as in the previous question, but this time we are revolving the region about a horizontal axis, so we slice vertically and integrate with respect to x. A vertical slice has height y = x2 , which when revolved produces a disk with radius r = x2 , so when the whole region is revolved, the resulting solid has

volume  = π \0 2  (x2 ) 2  dx = π \0 2 x4 dx = π [5(x5)]0(2)  = π (5(25) − 5(05)) = 32π5

A14.   Find \ 4x3 ln xdx.

E: x4 (xln x − x) + C

Solution:  We need integration by parts. Letting u = ln x and dv = 4x3 dx gives du =x(1) dx and v = x4 , so we see that

\ 4x3 lnxdx = (ln x)(x4 ) − \(x4 ) (   )x(1) dx = x4 lnx − \ x3 dx = x4 ln x − 4x4 + C

A15.   Evaluate \ xex dx.

3

A: 2 ln3 − 3 ln2 + 1     B: 3 ln3 − 2 ln2 + 1    C:    (ln 3)2 − (ln2)2

D: 2 ln3 − 3 ln2 − 1    E: 3 ln3 − 2 ln2 − 1

Solution:  We again need integration by parts.  Choosing u = x and dv = ex dx we have du = dx and v = ex , so we see that

\ xex dx = xex − \ ex dx = xex − ex + C = ex (x − 1) + C

Using ex (x − 1) as an antiderivative of xex , for the deinite integral we have

\ xex dx    =   [e  (xx − 1)] = eln 3 (ln3 − 1) − eln 2 (ln2 − 1) = 3(ln3 − 1) − 2(ln2 − 1)

=   3 ln3 − 3 − 2 ln2 + 2 = 3 ln3 − 2 ln2 − 1

A16.   Find \0 1  (x + 1 x − 2) dx.

Solution:  We need to ind the partial fraction decomposition of the integrand function:

3                    A            B          A(x − 2) + B(x + 1)

(x + 1)(x − 2)      x + 1     x − 2            (x + 1)(x − 2)

We see that we need A(x − 2) + B(x + 1) = 3. When x = − 1 we get A(−3) + B(0) = 3, so A = − 1, and

\0 1 dx   =   \0'(1)'