anonymous
  • anonymous
how to integrate x4^-xsquare?
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
plz tell me how to solve Q.29 and 30
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Zarkon
  • Zarkon
29 is a u-substitution
anonymous
  • anonymous
i do know that but how are we gonna do that?

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Zarkon
  • Zarkon
\[u=-x^2\]
anonymous
  • anonymous
but what about the 4?
anonymous
  • anonymous
plz elaborate a bit
Zarkon
  • Zarkon
\[\int a^xdx=\frac{a^x}{\ln(a)}+c\]
anonymous
  • anonymous
integration by parts wud be applied here?
Zarkon
  • Zarkon
no
anonymous
  • anonymous
then how m i gonna proceed? :( plz tell me ho to start it
anonymous
  • anonymous
how*
Zarkon
  • Zarkon
i told you \(u=-x^2\)
Zarkon
  • Zarkon
I'm talking about #29
anonymous
  • anonymous
that i know but 4 is also there it wud be x4^u * -du/2x
anonymous
  • anonymous
right?
anonymous
  • anonymous
\[u = -x^2 ---> du -2x dx\] \[-\frac{ 1 }{ 2 } \int\limits_{}^{}4^u du\]
anonymous
  • anonymous
but yes. what you wrote is right; simplified to what i wrote
anonymous
  • anonymous
ive reached this step but after that how should i solve m stuck there
anonymous
  • anonymous
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
anonymous
  • anonymous
u = -x^2**
anonymous
  • anonymous
okaaay thanks a bunch :)
anonymous
  • anonymous
glad i could help :)
anonymous
  • anonymous
can u plz help me with q30 as well?
anonymous
  • anonymous
yes. using the same rule: \[\int\limits_{}^{} (2^{\pi})^{x}dx = \frac{ (2^{\pi})^{x} }{ \ln(2^{\pi}) } + C = \frac{ (2^{\pi})^{x} }{ \pi \ln(2) } + C\]
anonymous
  • anonymous
okaaay thankyou
anonymous
  • anonymous
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 ??
anonymous
  • anonymous
rule which is demonstrated in #30: \[\log(a^b) = b \log(a)\] 2ln4 = ln16
anonymous
  • anonymous
okaay
anonymous
  • anonymous
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 \(u=x\ln a\), \(du=\ln a~dx\): \[\begin{align*}\int a^x~dx&=\int e^{u}~\frac{1}{\ln a}~du\\ &=\frac{1}{\ln a}\left(e^u+C\right)\\ &=\frac{1}{\ln a}e^u+C\\ &=\frac{1}{\ln a}e^{x\ln a}+C\\ &=\frac{a^x}{\ln a}+C \end{align*}\]

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