MATH SOLUTIONS 101 - PRACTICE MIDTERM I
Problem I. (12 pts) Short answer questions. Here you only need to give the correct answer no detailed explanations/ calculations are needed to support your answer. (a) (2 pts)
sin3 x cos2 x dx = 0
SOLUTION: Use symmetry: sin (x) = sin x, cos (x) = cos x, thus f (x) = f (x), where f (x) = sin3 x cos2 x is the integrand. Since the range of integration [, ] is symmetric to the origin the integral is 0. (b) (2 pts) Express the limit as a denite integral
1 n n lim
n i=1
n2 = n2 + i2
n2 n2 +i2
1 0
1 dx 1 + x2 set: xi = i/n and x = 1/n. Then the sum becomes:
SOLUTION: Write:
n i=1
=
1 , 1+(i/n)2
1 1 + x2 i
x
(c) (3 pts) Let R be the region between the curve y = sin 2x and the x-axis, for 0 x . Find 2 the area of the region. Simplify your answer completely! SOLUTION: If 0 x /2 then 0 2x , thus sin 2x 0. The area is:
/2
sin 2x dx =
0 2 0 f (x) dx
1 2
0
1 sin u du = cos u| = 1 0 2
(d) (3 pts) If
= 1 then nd
/4 0
f (2 tan ) sec2 d =
1 2
1
2
SOLUTIONS MATH 101 - PRACTICE MIDTERM I
SOLUTION: Let u = 2 tan , then du = 2 sec2 d. Also if x = 0 then u = 0, and if x = /4 then u = 2. Hence after this substitution the integral becomes one-half of the integral given in the problem. (e) (2 pts) Write down the n-th right Riemann approximating sum for the integral
n i=1
2 1
1 (x + ) dx = x
1 n
1+
i 1 + n 1+
i n
Full Answer questions. Here you have to give detailed explanations and/or calculations to support your answer. Problem II. (8 pts) Evaluate the following indenite integrals 1 2 1 1 (u 1)u1/2 = u5/2 u3/2 + C = 5 3
(a)
x3
x2 + 1 dx =
1 1 = (1 + x2 )5/2 (1 + x2 )3/2 + C 5 3 SOLUTION: Substitute u = 1 + x2 , then 1 du = x dx, and x2 = u 1, so x3 dx = 1 (u 1) du. 2 2
/4
(b)
0
tan3 (x) dx
SOLUTION: Use sec2 x = 1 + tan2 x and write tan3 (x) = tan x (sec2 x 1).
/4 0
tan3 (x) dx =
0
/4
tan x sec2 x dx
0
/4
sin x dx cos x the second term = ln (1/ 2) =
If u = tan x then du =
sec2 x,
so the rst terms becomes
substitute: u = cos x, du = sin x dx, so it becomes: 2/2. ln The answer is: 1/2 ln 2/2.
1 u du = 1/2. For 0 1/ 2 1/ 2 1 u du = ln u|1 1
Problem III. (6 pts) Let R be the region bounded by the curves y = x2 and y = 2x x2 . Set up an integration in the variable x for the following quantities: (a) The volume obtained by revolving R about the x-axis is obtained by using washers, The intersection points are: x2 = 2x x2 , so 2x = 2x2 , x = 0 or x = 1, thus
1
V=
0
((2x x2 )2 x4 ) dx
SOLUTIONS MATH 101 - PRACTICE MIDTERM I
3
(b) The volume obtained by revolving R about the y -axis obtained by using cylindrical shells: r = x, h = 2x x2 x2 . Thus
1
V=
0
2x (2x 2x2 ) dx
(c) The volume obtained by revolving R about the line x = 2 is again obtained by using cylindrical shells: r = 2 + x, h = 2x x2 x2 .
1
V=
0
2 (2 + x) (2x 2x2 ) dx
Problem IV. (4 pts) A solid has circular base of radius 1. Find the volume of the solid if the cross-sections perpendicular to the base are isosceles right triangles with hypotenuse in the base The hypotenuse which is perpendicular to the diagonal of the base circle going through the point x has length: l = 2 1 x2 by the Pythagorean theorem. The area of the x-cross section is: A(x) = l2 /4 = 1 x2 . The volume of the solid is: V= 1 4 1 x2 dx = x x3 |1 1 = 3 3 1
1
Problem V. (4 pts) Prove the following inequality. Explain all steps in your proof:
/4 0
1 sin (x2 ) dx 1 2
Since /4 < 1 then x2 x for 0 x /4 and then sin (x2 ) sin x. Then
/4 0
sin (x2 ) dx
0
/4
sin x dx = cos x |0
/4
1 =1 2
Problem VI. (6 pts) Find the area A of the region under the graph y = 2x x2 above the interval [0,2], by completing the following steps. a) (2 pts) The n-th approximating Riemann sum to the area is obtained by taking: xi = 2i/n, x = 2/n:
4
SOLUTIONS MATH 101 - PRACTICE MIDTERM I
Rn =
2 n
n i=1
4i 4i2 2 n n
b) (4 pts) Evaluate and simplify the sums you obtained and nd the limit as n . 8
n i=1 i n2
Rn =
8
n 2 i=1 i n3
=
4n(n + 1) 8n(n + 1)(2n + 1) = n2 6n3
= 4(1 +
1 4 1 1 ) (1 + )(2 + ) n 3 n n 8 4 = 3 3
thus
n
lim Rn = 4