1. Compute
/
|
| x sin(x) dx
|
/
Answer is d) -x cos(x) + sin(x) +C. This answer is obtained
by integration by parts by letting u = x and dv = sin(x).
The integration by parts formula is
/ /
| |
| u dv dx = uv - | v du dx
| |
/ /
2. Compute
1
/
| 2
| x exp(x ) dx
|
/
0
The answer is b) (e-1)/2. The antiderivative is (1/2)exp(x^2).
Inserting the limits 0..1 gives the answer.
3. Compute
infinity
/
| x
| --------- dx
| 2 2
/ (1 + x )
1
The answer is a) 1/4. The antiderivative is
1
- --------
2
2 + 2 x
This expression goes to zero as x -> infinity.
The other limit x=1 gives 1/4.
4. Set up the integral that computes the area between y = 2x+1 and y = 7+x-x^2.
The answer is: c)
2
/
| 2
| 6 - x - x dx
|
/
-3
Note that the parabola lies above the line over the interval [-3, 2]
> plot({2*x+1,7+x-x^2},x=-3..2);
5. Set up the integral required to compute the area between the curves y=x-1 and x+1=y^2.
The answer is e)
2
/
| 2
| y + 2 - y dy
|
/
-1
The straight line x = y+1 lies to the right of the parabola x = y^2-1
6. Set up the integral that computes the volume of the solid generated by revolving the reation between the curves y = 2x and y = x^2 about the y axis.
The answer is: a)
2
/
| 2
| 2 Pi x (2 x - x ) dx
|
/
0
The technique is cylindrical shells where the radius is x and the height
is 2x-x^2.
7. Compute
/
| 1
| ------- dx
| 3 2
/ x - x
The answer is c)
| -1 + x |
ln(----------) + 1/x + C
| x |
The partial fractions decompositition of the integrand is
1 1
- ---- - 1/x + ------
2 -1 + x
x
The integral of this expression is 1/x -ln|x| +ln|-1+x|.
The final answer is obtained by combining the difference of the logarithms
as the logarithm of the quotient.
8. Use the substitution x= 2 tan(t) to convert
2
/
| 1
| ----------- dx
| 2 1/2
/ (x + 4)
0
into an integral which just involves t.
The answer is: a)
1/4 Pi
/
|
| sec(t) dt
|
/
0
Note that x^2+4 = 4(tan^2 +1) = 4 (sec(t))^2.
Also dx = 2 (sec(t))^2 dt. So the integrand becomes sec(t).
Since tan(0)=0 and 2 tan(Pi/4)= 2, the t- limits become
0.. Pi/4.
9. The graph of a function f is given below. Which of the following statements most accurately describes the relationship between
2
/
|
| f(x) dx
|
/
0
and its approximation by T_4, the trapezoidal rule with 4 subdivisions.
The answer is b): T_4 is less than the integral. The reason is because the graph of the function is concave down and so the trapezoids lie below the graph (and therefore have less area).
10. Which statement most accurately describes the convergence or divergence of
infinity
/
| x
| ----------- dx
| 5 1/2
/ (x + 1)
1
The answer is e): The integral converges because
the integrand is comparable to x^(-3/2) by counting leading powers of x.
The integral of x^(-3/2) from 1 to infinity is a finite number (its value is 2) and so
the integral given in the problem is also finite.
11. Compute
/
| 3
| cos(x) dx
|
/
The answer is
3
sin(x) - 1/3 sin(x) + C
First write cos^3 as (1-sin^2) cos.
Then use u-substitution with u= sin(x) and du = cos(x) dx.
The integrand becomes 1-u^2 du, which when integrated
is u-u^3/3. Substituting u=sin(x) gives the answer.
12. Compute
/
| 2
| x exp(-x) dx
|
/
The solution is
2
- x exp(-x) - 2 x exp(-x) - 2 exp(-x) + C
Perform two integrations by parts. For the first, let
u=x^2 and dv = exp(-x) dx - - with du = 2x dx and v = - exp(-x).
The integration by parts formula is
/ /
| |
| u dv dx = uv - | v du dx
| |
/ /
The first integration by parts yields
/
2 |
-x exp(-x) + | 2 x exp(-x) dx
|
/
Then do another integration by parts with u = 2x and dv = exp(-x) dx
to obtain the final answer.
13. Compute
/
| 2 1/2
| (9 - x ) dx
|
/
The answer is
2 1/2
9/2 arcsin(1/3 x) + 1/2 x (9 - x ) + C
Make the trig substitution x=3sin(t) with dx = 3 cos(t) dt.
The integrand becomes 9 cos^2 , which can be integrated
with the identity (cos t)^2 = (1/2) (1+cos(2t)).
The t-integral becomes
(9/2)(t+sin(2t)/2) = (9/2)(t + cos(t) sin(t)).
Now we must convert back to x.
Recall x=3sin(t) and so sin(t) = x/3 and t = arcsin(x/3).
Also cos(t) = sqrt(9-x^2). Inserting these expressions, the
answer is obtained.
14. Set up the integral required to compute the volume of the region whose base is a unit circle in the $x$, $y$ plane and whose cross sections perpendicular to the $x$-axis are equilateral triangles. You do not need to compute this integral.
1
/
| 1/2 2
| 3 (1 - x ) dx
|
/
-1
which computes to 4 sqrt(3)/3. The key is to compute the area, A(x),
of the cross section at x. Then the answer is the integral of A(x)
from x=-1 to 1. Now the cross section is a triangle with base 2 sqrt(1-x^2)
(see figure). Since the triangle is an equilateral triangle, the height is
equal to sqrt(3)/2 times the length of the base or sqrt(3) sqrt(1-x^2).
The area of the triangle is (1/2) base X height, and so
A(x) = sqrt(3)(1-x^2).
15. The error in approximating the integral of f over a <= x <= b by using trapezoidal rule is no bigger than K (b-a)^3 / 12 n^2 where K is the maximum value of |f''| over the interval a <= x <= b and n is the number of subdivisions. Find the minimum number of subdivisions (n) that will be ensure that the error in computing
4 / | | ln(x) dx | / 2by trapezoidal rule is less than 1/400=0.0025.
The answer is n=9. Note that f'' is -1/x^2 which attains 1/4 as its maximum absolute value (at x=2). So the error expression becomes (1/4)*8/(12n^2) = 1/(6n^2). Setting this equal to 1/400 and solving for n gives n=20/sqrt(6)=8.16. The next integer larger is n=9.