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Unformatted text preview: Math 20B Second Midterm Solutions November 15, 2004 1. The curve goes from x = 0 to x = 1. Solving for y(x), we have y =
y ′ = −2x3 (1 − x4 )−1/2 .
1 1 1 + (y ′ )2 dx = (a) 1+
0 0 4x6
dx =
1 − x4 2πy(x) 1 + (y ′ )2 dx = 2π 0 (1 − x4 ) 1 + 0 0 1 = 2π
0 1 − x4 . Thus 1 − x4 + 4x6
dx.
1 − x4 1 1 (b) 1 √ 4x6
1 − x4 dx 1 − x4 + 4x6 dx. 2. (a) 0.08/42 = 0.005, since error is roughly C/n2 and we have multiplied n by 4.
(b) The error in the Midpoint Rule is about half that of the Trapezoidal Rule and
opposite in sign. Hence the answer is −0.04. (You will get partial credit for 0.04.)
√
√
3. (a) 2 cos(3π/4) + 2i sin(3π/4) = − 2 + 2 i.
(b) r = 21/3 . The values of θ are
3π/4
π
= ,
3
4 3π/4 + 2π
11π
=
,
3
12 3π/4 + 4π
19π
=
.
3
12 (You need not do the arithmetic to simplify the angles.)
4. Separating variables: dy = 2x dx and so ln y = x2 + C. Using y(0) = 2, we have
y
ln 2 = 02 + C and so C = ln 2. At x = 3 we have ln y = 32 + C = 9 + ln 2 and so
y = e9+ln 2 = 2e9 . 5. We have 2x2
2
1
1
= 2+
= 2+
−
,
2−1
x
(x − 1)(x + 1)
x−1 x+1 where it is up to you how you ﬁnd the partial fractions. Integrating gives 2x +
ln x−1 +C.
x+1 ...
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This note was uploaded on 04/29/2010 for the course MATH MATH 20B taught by Professor Takeda during the Spring '07 term at UCSD.
 Spring '07
 Takeda
 Math

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