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

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

Evaluate

$$\frac{4}{3}\approx 1.333333333$$

Solution Steps

Evaluate the indefinite integral first.

$$\int x^{2}+x^{2}\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{2}\mathrm{d}x+\int x^{2}\mathrm{d}x$$

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}.

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

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}.

$$\frac{x^{3}+x^{3}}{3}$$

Simplify.

$$\frac{2x^{3}}{3}$$

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.

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

Simplify.

$$\frac{4}{3}$$