Solve ∫ (from - 2 to 2) of 8-2x^2-4-x^2 wrt x

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

Evaluate

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Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

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

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}.

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

Find the integral of -4 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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 -1 times \frac{x^{3}}{3}.

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

Simplify.

$$4x-x^{3}$$

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.

$$4\times 2-2^{3}-\left(4\left(-2\right)-\left(-2\right)^{3}\right)$$

Simplify.

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