Question

Evaluate the definite integral.

Answer 1

Try "u" substitution

Answer 2

u=x^3
du=3x^2
\[\frac{ 1 }{ 3 } \int\limits_{0}^{3} e^u\]

Answer 3

*du=3x^2
\[\frac{ 1 }{ 3 } \int\limits_{0}^{2} e^u du\]

Answer 4

dx at the end of the 3x^2

Answer 5

Integrate that and then just plug the the values back in and finish it

Answer 6

what value?

Answer 7

x^3 and 3x^2?

Answer 8

no, just x^3

Answer 9

\(du = 3x^2 dx \Rightarrow \dfrac{1}{3}du = x^2 dx\)

Answer 10

\[Let I=\int\limits_{0}^{2}x ^{2}e ^{x ^{3}}dx\]
\[substitute x ^{3}=t\]
\[differentiate,3x ^{2}dx=dt, x ^{2}dx=\frac{ dt }{ 3 }\]
when x=0,\[t=0^{3}=0\]
whenx=2,\[t=2^{3}=8\]
\[I=\int\limits_{0}^{8}e ^{t}\frac{ dt }{ 3 }=\frac{ 1 }{3}e ^{t}from 0 \to 8\]
\[I= \frac{e ^{8}-e ^{0} }{ 3 }=\frac{ e ^{8}-1 }{3 }\]