Question

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

Answer 1

Question #50930, Math, Integral Calculus

Integrate with respect to $x$:

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

Solution:

a) If this $\int 2^{-1}x^2(x^3 + 4) \, dx$ means Integrate with respect to $x$:

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

Then

$$\begin{aligned} \int \frac{1}{2} x^2(x^3 + 4)^2 \, dx &= \frac{1}{2 \cdot 3} \int (x^3 + 4)^2 d(x^3) = \frac{1}{6} \int (x^3 + 4)^2 d(x^3 + 4) \\ &= \frac{(x^3 + 4)^3}{6 \cdot 3} + C \end{aligned}$$

Where $C$ is arbitrary constant. Here we use that $\int x^\alpha \, dx = \frac{x^{\alpha+1}}{\alpha+1} + C$ for all $\alpha \neq -1$.

b) If this $\int 2^{-1}x^2(x^3 + 4) \, dx$ means Integrate with respect to $x$:

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

Then

$$\begin{aligned} \int_{-1}^{2} x^2(x^3 + 4)^2 \, dx &= \frac{1}{3} \int_{-1}^{2} (x^3 + 4)^2 d(x^3) = \frac{1}{3} \int_{-1}^{2} (x^3 + 4)^2 d(x^3 + 4) = \frac{(x^3 + 4)^3}{9} \Bigg|_{-1}^{2} \\ &= \frac{(2^3 + 4)^3}{9} - \frac{(-1)^3 + 4)^3}{9} = \frac{12^3}{9} - \frac{3^3}{9} = 24 - 3 = 21 \end{aligned}$$

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