Question

use integration by parts: integral of 2^(sqrtx)

Answer 1

\[\int\limits_{4}^{9} 2^{\sqrt{x}} dx\]
\[u= 2^{\sqrt{x}}\] \[du= 2^{\sqrt{x}} (\ln2)(\frac{ 1 }{ 2\sqrt{x}})\]
\[dv= 1\] \[v= x\]
\[= x(2^{\sqrt{x}}) - \frac{ \ln2 }{ 2 } \int\limits 2^{\sqrt{x}}(\sqrt{x})\]

Answer 2

is the substitution given ?

Answer 3

no, but i have to use Integration by parts..

Answer 4

\[u= 2^{\sqrt{x}}\] \[du= 2^{x}(\ln2)(\frac{ 1 }{ 2x }) dx\]\[dv= \sqrt{x}\] \[v=\frac{ 2 x ^{\frac{ 3 }{ 3 }}}{ 3 }\]
is this substitution right?

Answer 5

\[v ={\frac{2x^{\frac{ 3 }{ 2 }}}{3}} \]

Answer 6

This is a tricky one. I know the answer but I don't know how it's done.
give me 24 hours

Answer 7

i'm trying to solve it but i'm still writing it in LaTex to post here

Answer 8

d/dx a^x = a ln a

Answer 9

a^x ln a*

Answer 10

Manipulate a little bit
let u = 2^(sqrt x) --> \(log_2 u = \sqrt x\)--> x = (\(\log_2 u)^2 \)

Answer 11

try this way you can get the answer :)

Answer 12

sorry i don't get it.
but thanks :)

Answer 13

§ 2\(\normalsize\color{slate}{ ^{\sqrt{x}} }\) dx

u= \(\normalsize\color{slate}{ \sqrt{x} }\) du=1 / 2\(\normalsize\color{slate}{ \sqrt{x} }\)

2§ 2\(\normalsize\color{slate}{ ^{u} }\) u du


f=u
ds=2\(\normalsize\color{slate}{ ^{u} }\) du
df=du
s= 2\(\normalsize\color{slate}{ ^{u} }\) / log(2)


2§ 2\(\normalsize\color{slate}{ ^{u+1} }\) u 2§ 2\(\normalsize\color{slate}{ ^{u+1} }\)
--------- - ---------- +C
log(2) log(2)

Answer 14

Damn these question marks... os always has issues -:(

Answer 15

the integral signs shouldn't be there in a last line, my bad

Answer 16

this is what it should be, without 2§ in the begging of each fraction.