Solve ∫ (from - 1 to 1) of [(x^2-1)*(-x^2+1)] wrt x

$$\int _ { - 1 } ^ { 1 } [ ( x ^ { 2 } - 1 ) \cdot ( - x ^ { 2 } + 1 ) ] d x$$

Evaluate

$$-\frac{16}{15}\approx -1.066666667$$

Solution Steps

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

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

Multiply -1 and -1 to get 1.

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

Combine x^{2} and x^{2} to get 2x^{2}.

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

To multiply powers of the same base, add their exponents. Add 2 and 2 to get 4.

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

Evaluate the indefinite integral first.

$$\int -x^{4}+2x^{2}-1\mathrm{d}x$$

Integrate the sum term by term.

$$\int -x^{4}\mathrm{d}x+\int 2x^{2}\mathrm{d}x+\int -1\mathrm{d}x$$

Factor out the constant in each of the terms.

$$-\int x^{4}\mathrm{d}x+2\int x^{2}\mathrm{d}x+\int -1\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{4}\mathrm{d}x with \frac{x^{5}}{5}. Multiply -1 times \frac{x^{5}}{5}.

$$-\frac{x^{5}}{5}+2\int x^{2}\mathrm{d}x+\int -1\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}. Multiply 2 times \frac{x^{3}}{3}.

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

Find the integral of -1 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$-\frac{x^{5}}{5}+\frac{2x^{3}}{3}-x$$

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.

$$-\frac{1^{5}}{5}+\frac{2}{3}\times 1^{3}-1-\left(-\frac{\left(-1\right)^{5}}{5}+\frac{2}{3}\left(-1\right)^{3}-\left(-1\right)\right)$$

Simplify.

$$-\frac{16}{15}$$