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The solutions to exam i for math106b calculus ii, taught by professor p. Wong, held on february 3, 2006. The exam covers topics such as finding the exact area of a region bounded by two curves, calculating the arc-length of a path, evaluating definite and indefinite integrals, and determining the volume of a solid of revolution. Problems with detailed solutions and explanations.
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EXAM I - FEBRUARY 3, 2006
Instruction: Read each question carefully. Explain ALL your work and give reasons to support your answers. Advice: DONโT spend too much time on a single problem.
Problems Maximum Score Your Score
1
2 EXAM I - FEBRUARY 3, 2006
1.(16 pts.)(a) Consider the region bounded by the graph of y = 2x โ x^2 and the graph of y = x^4. Find the exact area of the region. [Make a sketch of the region first.]
y=2xโx^2
As shown in the figure above. The two curves intersect at two points, one at the origin and the other at (1, 1). On a typical slice, the curve y = 2x โ x^2 is above the curve y = x^4. Therefore, the area of the bounded region is given by
0
(2x โ x^2 ) โ x^4 dx
= x^2 โ x
3 3 โ^
x^5 5
1 0 = (1 โ 13 โ 15 ) โ 0 = 157.
(4 pts.)(b) Write (DO NOT evaluate) a definite integral representing the arc-length of the path given by y = 2x โ x^2 from the origin (0,0) to the point (1,1). Since y = 2x โ x^2 , yโฒ^ = 2 โ 2 x and (yโฒ)^2 = 4 โ 8 x + 4x^2. The length of the path from (0, 0) to (1, 1) is given by
L =
0
1 + (4 โ 8 x + 4x^2 ) dx =
0
5 โ 8 x + 4x^2 dx.
4 EXAM I - FEBRUARY 3, 2006
3.(20 pts.) The region bounded by the graph of y = โx, the x-axis, and the line x = 2 is revolved about the y-axis. Find the exact volume of the resulting solid of revolution. [Sketch a picture of the region and the solid.]
y
As shown in the figure, a typical slice is a washer of thickness โy. When x = 2, the point on the curve of y = โx has y-coordinate equal to
V =
0
ฯ(2^2 โ (y^2 )^2 ) dy
= ฯ
0
4 โ y^4 dy
= ฯ
4 y โ y
5 5
2 ฯ
2 ฯ
MATH106B CALCULUS II - PROF. P. WONG 5 4.(10 pts.)(a) Evaluate the indefinite integral โซ (^2) x (^3) โ x 2 2 x^2 โ x โ 3 dx. First use long division to obtain that 2 x^3 โ x^2 = x(2x^2 โ x โ 3) + 3x. Thus, (^) โซ 2 x^3 โ x^2 2 x^2 โ x โ 3 dx^ =
x + (^2) x (^2) โ^3 x x โ 3 dx
= x
2 2 +
โซ (^3) x (x + 1)(2x โ 3) dx. Using the method of partial fraction, we write 3 x (x + 1)(2x โ 3) =^
x + 1 +^
2 x โ 3 and so 3 x โก A(2x โ 3) + B(x + 1).
By evaluating the left hand side at x = โ 1 and at x = 32 , we find that A = 35 and B = 95. Now, โซ (^2) x (^3) โ x 2 2 x^2 โ x โ 3 dx^ =^
x^2 2 +
โซ (^) dx x + 1 +
โซ (^) dx 2 x โ 3 = x
2 2 +
5 ln^ |x^ + 1|^ +
10 ln^ |^2 x^ โ^3 |^ +^ C. (10 pts.)(b) Evaluate the definite integral โซ (^) e 1
ln x x^2 dx.
Let u = ln x and dv = xโ^2 dx. Thus, du = dx x and v = โ (^1) x. It follows from the technique of integration by parts that โซ (^) e 1
ln x x^2 dx^ =^ โ^
ln x x
e 1
โซ (^) e 1
โ (^) x^1
ยท (^1) x dx
= โ ln x^ x
e 1
โซ (^) e 1
xโ^2 dx
= โ ln x^ x
e 1
โ 1 โ 1
e 1 = โ (^) x^1 โ ln x^ x
e 1
= 1 โ (^2) e.
