# Evaluating a definite integral with e

Consider the following problem, evaluate:

$$\int_0^1 \dfrac{x^2}{e^{2x^3}} dx$$

I tried the following method:

$\ln(x) = u$

$$\dfrac{1}{e^u} \cdot \dfrac{1}{6}du$$

$$\dfrac{1}{6} \cdot \ln(e^{2x^3}) + c$$

But if I use this I get incorrect answer. Where did I go wrong and what would I have to do to make correct use of this method to solve the problem?

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I'm not quite certain what you did - usually when you substitute you are looking for something in the integrand to substitute out (or more advanced, to reduce the integrand into something "nicer", which can lead to some creative substitutions).

A good rule of thumb for intro calculus is, look for something whose derivative is already in the problem or almost in the problem, and pick that for your u. Here, I would make the substitution $u=2x^3$ since $u'=6x^2$ differs only by a constant from $x^2$.

Your integral becomes $\frac{1}{6}\int_0^2 e^{-u} du$. Can you work out the rest?

NB to OP (taking into consideration some of the other answers): Remember you're doing a definite integral so you shouldn't have $+C$ in your answer. Some of the other answers that have it aren't wrong but they've only taken you up to the antiderivative step; you still have to evaluate it at your bounds of integration.

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Use $u=x^3$, then $dx = \frac{du}{3x^2}$. This turns the integral into: $$\frac{1}{3}\int\limits_{0}^{1}e^{-2u}du = \frac{-1}{6}e^{-2u}+C=\frac{-1}{6}e^{-2x^3} +C$$

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$$u=2x^3$$ $$du=6x^2\space{dx}$$ $$\int\frac{x^2}{e^{2x^3}}dx$$ $$\frac{1}{6}\int{e^{-u}}du$$ $$-\frac{1}{6}e^{-u}+C=-\frac{1}{6}e^{-2x^3}+C$$

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So basically only my very last step was wrong? – MartinBurg May 27 '13 at 19:00
@MartinBurg You made $u=\ln{x}$, and said the integrand was equal to $\frac{1}{e^u}\frac{1}{6}$, which is incorrect given your choice of $u$. – Ataraxia May 27 '13 at 19:13

$$\int_0^1 \dfrac{x^2}{e^{2x^3}} dx$$

Let $u=2x^3$. Then $du = 6x^2 dx$, and we have:

$$\frac16 \int_0^1 \dfrac{6x^2}{e^{2x^3}} dx$$ $$\frac16 \int_0^2 \dfrac{1}{e^u} du \tag{Limits change}$$ $$\frac16 \int_0^2 \dfrac{1}{e^u} du$$

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So only my last step was wrong, am I right? Since my second to last step is equal to your last step (apart from some sloppy writing on my part)? – MartinBurg May 27 '13 at 19:01

Do this:

$u= 2x^3$ , then $du = 6x^2 dx$ . when $x = 0$ we have $u = 0$. When $x=1$ we have $u=2$. Then

$$\int_{0}^{1} \frac{x^2}{e^{2x^3}} dx = \frac{1}{6}\int_{0}^{1} 6 \frac{x^2}{e^{2x^3}} dx$$

$$\int_{0}^{2} \frac{1}{e^u} du = \int_{0}^{2}e^{-u} du = - e^{-2} + e^{-0}$$

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