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

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

Evaluate

$$\frac{28}{3}\approx 9.333333333$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

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

$$\frac{2x^{3}}{3}+4x$$

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.

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

Simplify.

$$\frac{28}{3}$$