Solve ∫ (from 0 to 1) of (x^2+4)(x^2-2) wrt x

$$\int _ { 0 } ^ { 1 } ( x ^ { 2 } + 4 ) ( x ^ { 2 } - 2 ) d x$$

Evaluate

$$-\frac{107}{15}\approx -7.133333333$$

Solution Steps

Use the distributive property to multiply x^{2}+4 by x^{2}-2 and combine like terms.

$$\int _{0}^{1}x^{4}+2x^{2}-8\mathrm{d}x$$

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

$$\frac{x^{5}}{5}+2\int x^{2}\mathrm{d}x+\int -8\mathrm{d}x$$

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}.

$$\frac{x^{5}}{5}+\frac{2x^{3}}{3}+\int -8\mathrm{d}x$$

Find the integral of -8 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{5}}{5}+\frac{2x^{3}}{3}-8x$$

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.

$$\frac{1^{5}}{5}+\frac{2}{3}\times 1^{3}-8-\left(\frac{0^{5}}{5}+\frac{2}{3}\times 0^{3}-8\times 0\right)$$

Simplify.

$$-\frac{107}{15}$$