Question

Evaluate the indefinite integral.

\(\)\(\int {{x^2}} {e^{{x^3}}}dx\)

Short answer

The value of \(\int {{x^2}} {e^{{x^3}}}dx\)is\(\frac{1}{3}{e^{{x^3}}} + C\).

Step-by-step solution

Definition of the indefinite integral

An integral that has no upper bound and no lower bound is known as indefinite integral.

Evaluating the integral.

The given data is written as,

\(\int {{x^2}} {e^{{x^3}}}dx = \int {{e^{{x^3}}}} \left( {{x^2}dx} \right)\) …(1)

Let \({x^3} = u\)

After differentiating the integral, we get

\(\begin{aligned}{c}3{x^2}dx = du\\{x^2}dx = \frac{1}{3}du\end{aligned}\)

Substitute all the value in the equation (1).

\(\begin{aligned}{c}\int {{x^2}} {e^{{x^3}}}dx = \int {{e^u}} \left( {\frac{1}{3}du} \right)\\ = \frac{1}{3}\int {{e^u}} du\end{aligned}\)

Since the derivative of \({e^x}\)is\({e^x}\)itself. The anti-derivative of\({e^x}\)is also\({e^x}\)

\( = \frac{1}{3}{e^u} + C\)

Substitute again, we get

\(\begin{aligned}{c}u = {x^3}\\ = \frac{1}{3}{e^{{x^3}}} + C\end{aligned}\)

Hence The value of \(\int {{x^2}} {e^{{x^3}}}dx\)is\(\frac{1}{3}{e^{{x^3}}} + C\).