# Thread: Help with integral with upper/lower limts

1. ## Help with integral with upper/lower limts

Q. Evaluate $\displaystyle \int \frac{6}{x^2}dx$

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

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.

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

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

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:

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

RonL

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

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

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

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

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

$\displaystyle \int_{4}^{1} \frac {(4+\sqrt x)^2}{2\sqrt x}dx$$\displaystyle = \left[ \frac{1}{3} \sqrt x(x+12 \sqrt x +48) \right]_{4}^{1}$
$\displaystyle =(1+12+48)/3 -(4+12.2+48)2/3$

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