Question

how to evaluate the integrals
x/(3+sqrt(x))
improper integral
from -x to 0 e^xdx

Answer 1

\[\int\limits_{}^{} \frac{ x }{ 3+\sqrt x }dx\]

Answer 2

is this your question?

Answer 3

yes

Answer 4

i let U=\[\sqrt{x}\]
U^2 = x
2Udu=dx

Answer 5

Make u=x^2

Answer 6

du = 2 x dx

Answer 7

then im stuck at \[\int\limits_{?}^{?} 2u^3/(3+u)du\]

Answer 8

oh ok let me try that

Answer 9

I am changing my mind. Wait a minute

Answer 10

sorry but im stuck at \[\sqrt{x}\] when substituting in the u value in

Answer 11

Actually, your first substitution works better, I think

Answer 12

Notice that
\[
\frac{2 u^3}{u+3}=2 u^2-6 u-\frac{54}{u+3}+18
\]

Answer 13

By doing long division. The rest is easy

Answer 14

thank you i get it now

Answer 15

Your final answer should look like
\[
\frac{2 x^{3/2}}{3}-3 x+18 \sqrt{x}-54 \log \left(\sqrt{x}+3\right)
\]

Answer 16

YW

Answer 17

can you help me on the improper integral

Answer 18

\[\int\limits_{-\infty}^{0} e^xdx\]

Answer 19

Use this calculator to understand the problem you stuck at NOT for solving assignments or something like that and remember that it's all about understanding
Good luck ;)
https://www.symbolab.com/solver/integral-calculator

Answer 20

do u know what is e^x integration?

Answer 21

if it was indefinite integral it's simply \[e^x+c\]
if it was definite integral it's simply\[\int\limits_{Lower } ^ {Upper} e^x dx =e^{Upper} - e^{Lower} \]

Answer 22

@zaxoanl:

\[\int\limits\limits_{-\infty}^{0} e^xdx\] ...
1) is an improper integral due to the negative infinity lower limit.
2) is a definite integral. See @HossamHoussien 's "definite integral" equation, above.

Applying this definite integral formula to your \[\int\limits\limits\limits_{-\infty}^{0} e^xdx, \]


\[\int\limits\limits\limits_{-\infty}^{0} e^xdx\]

Answer 23

\[\int\limits\limits\limits\limits_{-\infty}^{0} e^xdx=e^0-limit.as.x \rightarrow negative.infinity.of.e^x\]

Answer 24

Can you finish evaluating this definite integral?

Answer 25

\[e^0-e ^{-\infty }=1-\frac{ 1 }{ e^\infty }=1-0=1\]