Please help me find my calc mistake

Test Preparation
1

<p>What am I doing wrong on these problems? </p>

<li><p>Calculate the integral of x/(x^2 -3x +2)dx. I use partial fractions, and get the integral of -1/(x-1) dx + the integral of 2/(x-2) dx, which gives me the answer of 2ln[x-2] - ln[x-1] +C. The correct answer is ln[x-2] - ln[x-1] +C. </p></li>
<li><p>The arc y = x^2/2 from (2,2) to (4,8) is rotated around the y axis. Find the surface area. I get SA = 2pi * integral from 2 to 8 of sqrrt(2y) * sqrrt(1 + 1/(2y)) dy ----> 2pi * integral from 2 to 8 of sqrrt(2y +1)dy, which equals 2pi *(1/3)(2y+1)^(3/2), evaluated from 2 to 8. I get 2pi/3 *(17^1.5 -5^1.5). The answer is 2pi *(17^1.5 -5^1.5). </p></li>
<li><p>A solid has a circular base of radius 1. Parallel cross-sections perpendicular to the base are squares. The volume of the solid is? I get the integral from -1 to 1 of 4 *(1-x^2)dx —> 4x - (4x^3)/3, evaluated from -1 to 1, which equals 16/3. The answer is 4/3. </p></li>
</ol>

<p>Thanks a bunch!!!</p>

2

<p>1) So you got</p>

<p>x/(x-2)(x-1) = A/(x-2) + B/(x-1)</p>

<p>Multiplying by LCD, you get</p>

<p>x = A(x-1) + B(x-2)</p>

<p>When x=2, A=2
When x=1, B=-1</p>

<p>That leaves you with </p>

<p><integ> 2/(x-2) - 1/(x-1) dx</integ></p>

<p>= 2ln|x-2| - ln|x-1| + C</p>

3

<p>oops ignore that. hah.</p>

4

<ol>
<li>One way to verify your answer for any indefinite integral problem is to differentiate your answer, and see if you get the original expression.</li>
</ol>

<p>d( 2ln[x-2] - ln[x-1] +C) /dx
is indeed 2/(x-2) - 1/(x-1) or x/(x^2 -3x +2), so your answer is correct, not the book's.</p>

5

<p>thanks!! That makes me a little less nervous... any more comments on # 2, 3??</p>