Help with integral with upper/lower limts

• Aug 26th 2006, 09:35 AM
Yogi_Bear_79
Help with integral with upper/lower limts
Q. Evaluate $\int \frac{6}{x^2}dx$

A. $\int \frac{6}{x^2}dx=-\frac{6}{x}=???$

The evaluation equation has an upper limit of 3 and a lower limit of 1. I couldn't figure out how to show that with LaTEX. As you can see by my answer above, I have it partially completed. I'm stuck on how to finish it.
• Aug 26th 2006, 09:55 AM
CaptainBlack
Quote:

Originally Posted by Yogi_Bear_79
Q. Evaluate $\int \frac{6}{x^2}dx$

A. $\int \frac{6}{x^2}dx=-\frac{6}{x}=???$

The evaluation equation has an upper limit of 3 and a lower limit of 1. I couldn't figure out how to show that with LaTEX. As you can see by my answer above, I have it partially completed. I'm stuck on how to finish it.

Click on the first term to see how to set the limits:

$\int_{1}^{3} \frac{6}{x^2}dx$ $
=\left[-\frac{6}{x} \right]_1^3
$
$
=(-6/3)-(-6/1)=-2+6=4
$

RonL
• Aug 26th 2006, 10:48 AM
Yogi_Bear_79
To see if I understand, I have posted a second question that I worked brom begining to end, please validate.

Q. Evaluate $\int_{4}^{1} \frac {(4+\sqrt x)^2}{2\sqrt x}dx$

A. $\int_{4}^{1} \frac {(4+\sqrt x)^2}{2\sqrt x}dx = \left{ \frac{1}{3} \sqrt x(x+12 \sqrt x +48 \right]_{4}^{1}=51 -20 = 30$
• Aug 26th 2006, 11:08 AM
CaptainBlack
Quote:

Originally Posted by Yogi_Bear_79
To see if I understand, I have posted a second question that I worked brom begining to end, please validate.

Q. Evaluate $\int_{4}^{1} \frac {(4+\sqrt x)^2}{2\sqrt x}dx$

A. $\int_{4}^{1} \frac {(4+\sqrt x)^2}{2\sqrt x}dx = \left{ \frac{1}{3} \sqrt x(x+12 \sqrt x +48 \right]_{4}^{1}=51 -20 = 30$

First you TeX needs tidying up

$\int_{4}^{1} \frac {(4+\sqrt x)^2}{2\sqrt x}dx$ $
= \left[ \frac{1}{3} \sqrt x(x+12 \sqrt x +48) \right]_{4}^{1}$

$
=(1+12+48)/3 -(4+12.2+48)2/3$

$=61/3-152/3=91/3$
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