Question

Integrate with respect to x :
∫2−1x2(x3+4)2dx

Answer 1

Answer on Question #45966 – Math - Integral Calculus

Integrate with respect to $x$ :

$$\int_{2-1}^{2} x^2(x^3 + 4) \, dx$$

Solution:

$$\int_{-1}^{2} \frac{x^2}{(x^3 + 4)^2} \, dx$$

Now on to a friendly u-substitution:

Let $u = x^3 + 4$ then $\mathrm{du} = 3x^2 \, dx$, but since we need $x^2 \, dx$, it's $\frac{1}{3} \, \mathrm{du} = x^2 \, dx$.

The boundaries from $x = -1$ to 2 become $u = (-1)^3 + 4$ to $u = 2^3 + 4 \rightarrow u = 3$ to 12.

The integral in terms of $u$ is now:

$$\int_{3}^{12} \frac{1}{3} \frac{\mathrm{du}}{\mathrm{u}^2} = \frac{1}{3} \int_{3}^{12} \mathrm{u}^{-2} \, \mathrm{du} = -\frac{1}{3} \mathrm{u}^{-1} \left| \frac{12}{3} \right. = -\frac{1}{3} \left( \frac{1}{12} - \frac{1}{3} \right) = 1/12$$

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