Question

how to integrate x4^-xsquare?

Answer 1

plz tell me how to solve Q.29 and 30

Answer 2

29 is a u-substitution

Answer 3

i do know that but how are we gonna do that?

Answer 4

\[u=-x^2\]

Answer 5

but what about the 4?

Answer 6

plz elaborate a bit

Answer 7

\[\int a^xdx=\frac{a^x}{\ln(a)}+c\]

Answer 8

integration by parts wud be applied here?

Answer 9

no

Answer 10

then how m i gonna proceed? :( plz tell me ho to start it

Answer 11

how*

Answer 12

i told you \(u=-x^2\)

Answer 13

I'm talking about #29

Answer 14

that i know but 4 is also there
it wud be x4^u * -du/2x

Answer 15

right?

Answer 16

\[u = -x^2 ---> du -2x dx\]

\[-\frac{ 1 }{ 2 } \int\limits_{}^{}4^u du\]

Answer 17

but yes. what you wrote is right; simplified to what i wrote

Answer 18

ive reached this step but after that how should i solve m stuck there

Answer 19

because of the rule above for integrating a^x we have:

\[-\frac{ 1 }{ 2 }( \frac{ 4^u }{ \ln(4) }) + C\]

then you plug u = x^2 back in

Answer 20

u = -x^2**

Answer 21

okaaay thanks a bunch :)

Answer 22

glad i could help :)

Answer 23

can u plz help me with q30 as well?

Answer 24

yes.
using the same rule:

\[\int\limits_{}^{} (2^{\pi})^{x}dx = \frac{ (2^{\pi})^{x} }{ \ln(2^{\pi}) } + C = \frac{ (2^{\pi})^{x} }{ \pi \ln(2) } + C\]

Answer 25

okaaay thankyou

Answer 26

i got this answer of q,29
-1/2 4^-x square/ln(4) +c

the answer which in my book is
-4^-x square/ln 16+c

so do i have do do this to make it ln 16
(ln4)^2 ??

Answer 27

rule which is demonstrated in #30:

\[\log(a^b) = b \log(a)\]

2ln4 = ln16

Answer 28

okaay

Answer 29

If you're not familiar with the formula for integrating \(a^x\), \(a\not=0\):
\[\begin{align*}\int a^x~dx&=\int e^{\ln{a^x}}~dx\\
&=\int e^{x\ln a}~dx
\end{align*}\]
Make the sub \(u=x\ln a\), \(du=\ln a~dx\):
\[\begin{align*}\int a^x~dx&=\int e^{u}~\frac{1}{\ln a}~du\\
&=\frac{1}{\ln a}\left(e^u+C\right)\\
&=\frac{1}{\ln a}e^u+C\\
&=\frac{1}{\ln a}e^{x\ln a}+C\\
&=\frac{a^x}{\ln a}+C
\end{align*}\]