Question

consider the integral
\[\int\limits_{0}^{8} (x+e^x)^2 dx \]
a. use the midpoint rule with n=4 to approximate this integral.

Answer 1

HELP! :(

Answer 2

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

Answer 3

not really helping....

Answer 4

i have to use the midpoint rule not just only integrate

Answer 5

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
\[\Delta=(8-0)/4 =2 \]

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

thanks nadeem! big help :D

Answer 10

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

Answer 11

sure thing

Answer 12

\[\int\limits_{0}^{1} \ln(x) / x ^{1/2} dx \]

to approximate this integral using the trapezoid rule.
can you also help me with this :( sorryy

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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