## anonymous 5 years ago consider the integral $\int\limits_{0}^{8} (x+e^x)^2 dx$ a. use the midpoint rule with n=4 to approximate this integral.

1. anonymous

HELP! :(

2. anonymous

Expand the square, to get x^2+e^2x+2xe^x and integrate separately

3. anonymous

not really helping....

4. anonymous

i have to use the midpoint rule not just only integrate

5. anonymous

yes, so with the midpoint rule, you are asked to divide the curve into four rectangles. which means that $\Delta = (b-a)/n$ where b = 8 and a = 0 and n =4 so the width of the sub intervals is $\Delta=(8-0)/4 =2$

6. anonymous

$\Delta x=\frac {b-a}{n}\rightarrow \frac {8-0}{4}=2$ your intervals are:$[0,2],[2,4],[4,6],[6,8]$

7. anonymous

So your subintervals are (0.2), (2,4),(4,6) and (6,8) which means your mid points are at x =1,3,5,7

8. anonymous

so evaluate at the midpoints for each of the subintervals: $x=1,3,5,7$

9. anonymous

10. anonymous

therefore according to the rectangle rule, $\int\limits_{0}^{8}f(x)dx \approx 2 * \sum_{1}^{7}f(x)$ where x takes values 1,3,5,7

11. anonymous

sure thing

12. anonymous

$\int\limits_{0}^{1} \ln(x) / x ^{1/2} dx$ to approximate this integral using the trapezoid rule. can you also help me with this :( sorryy

13. anonymous

$\frac{\Delta x}{2}=\frac{\frac{b-a}{n}}{2}$ how many subintervals do you need? you didn't specify Here is the generic equation: $\int\limits\limits_{0}^{1} \frac{lnx}{\sqrt{x}}dx=\frac {\Delta x}{2}[\frac{lnx_0}{\sqrt{x_0}}+2*\frac {lnx_1}{\sqrt{x_1}}+...+2*\frac{lnx_{n-1}}{\sqrt{x_{n-1}}}+\frac {lnx_n}{\sqrt{x_n}}]$

14. anonymous

the question is determine if the improper integral is convergent or divergent. if its convergent, find its value.

15. anonymous

Here is the problem..... you did posted earlier

16. anonymous

yeah sorry and i got this already thanks, but can you help me with other one i posted on the previous section?