Solve ∫ (from -1 to 2) of left(x^2-1right)^3x wrt x

$$\int_{ -1 }^{ 2 } { \left( { x }^{ 2 } -1 \right) }^{ 3 } x d x$$

Evaluate

$$\frac{81}{8}=10.125$$

Solution Steps

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(x^{2}-1\right)^{3}.

$$\int _{-1}^{2}\left(\left(x^{2}\right)^{3}-3\left(x^{2}\right)^{2}+3x^{2}-1\right)x\mathrm{d}x$$

To raise a power to another power, multiply the exponents. Multiply 2 and 3 to get 6.

$$\int _{-1}^{2}\left(x^{6}-3\left(x^{2}\right)^{2}+3x^{2}-1\right)x\mathrm{d}x$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$\int _{-1}^{2}\left(x^{6}-3x^{4}+3x^{2}-1\right)x\mathrm{d}x$$

Use the distributive property to multiply x^{6}-3x^{4}+3x^{2}-1 by x.

$$\int _{-1}^{2}x^{7}-3x^{5}+3x^{3}-x\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int x^{7}-3x^{5}+3x^{3}-x\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{7}\mathrm{d}x+\int -3x^{5}\mathrm{d}x+\int 3x^{3}\mathrm{d}x+\int -x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{7}\mathrm{d}x-3\int x^{5}\mathrm{d}x+3\int x^{3}\mathrm{d}x-\int x\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{7}\mathrm{d}x with \frac{x^{8}}{8}.

$$\frac{x^{8}}{8}-3\int x^{5}\mathrm{d}x+3\int x^{3}\mathrm{d}x-\int x\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{5}\mathrm{d}x with \frac{x^{6}}{6}. Multiply -3 times \frac{x^{6}}{6}.

$$\frac{x^{8}}{8}-\frac{x^{6}}{2}+3\int x^{3}\mathrm{d}x-\int x\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{3}\mathrm{d}x with \frac{x^{4}}{4}. Multiply 3 times \frac{x^{4}}{4}.

$$\frac{x^{8}}{8}-\frac{x^{6}}{2}+\frac{3x^{4}}{4}-\int x\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x\mathrm{d}x with \frac{x^{2}}{2}. Multiply -1 times \frac{x^{2}}{2}.

$$\frac{x^{8}}{8}-\frac{x^{6}}{2}+\frac{3x^{4}}{4}-\frac{x^{2}}{2}$$

Simplify.

$$-\frac{x^{2}}{2}+\frac{3x^{4}}{4}-\frac{x^{6}}{2}+\frac{x^{8}}{8}$$

The definite integral is the antiderivative of the expression evaluated at the upper limit of integration minus the antiderivative evaluated at the lower limit of integration.

$$-\frac{2^{2}}{2}+\frac{3}{4}\times 2^{4}-\frac{2^{6}}{2}+\frac{2^{8}}{8}-\left(-\frac{\left(-1\right)^{2}}{2}+\frac{3}{4}\left(-1\right)^{4}-\frac{\left(-1\right)^{6}}{2}+\frac{\left(-1\right)^{8}}{8}\right)$$

Simplify.

$$\frac{81}{8}$$