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

$$\int_{ -2 }^{ 2 } 2 { y }^{ 2 } +4y d y$$

Evaluate

$$\frac{32}{3}\approx 10.666666667$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

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

Since \int y^{k}\mathrm{d}y=\frac{y^{k+1}}{k+1} for k\neq -1, replace \int y\mathrm{d}y with \frac{y^{2}}{2}. Multiply 4 times \frac{y^{2}}{2}.

$$\frac{2y^{3}}{3}+2y^{2}$$

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.

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

Simplify.

$$\frac{32}{3}$$