Solve ∫ (from -4 to 2) of left(2xright)^2+4x-16 wrt x

$$\int_{ -4 }^{ 2 } { \left(2x \right) }^{ 2 } +4x-16 d x$$

Evaluate

$$-24$$

Solution Steps

Expand \left(2x\right)^{2}.

$$\int _{-4}^{2}2^{2}x^{2}+4x-16\mathrm{d}x$$

Calculate 2 to the power of 2 and get 4.

$$\int _{-4}^{2}4x^{2}+4x-16\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 4x^{2}+4x-16\mathrm{d}x$$

Integrate the sum term by term.

$$\int 4x^{2}\mathrm{d}x+\int 4x\mathrm{d}x+\int -16\mathrm{d}x$$

Factor out the constant in each of the terms.

$$4\int x^{2}\mathrm{d}x+4\int x\mathrm{d}x+\int -16\mathrm{d}x$$

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 4 times \frac{x^{3}}{3}.

$$\frac{4x^{3}}{3}+4\int x\mathrm{d}x+\int -16\mathrm{d}x$$

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

$$\frac{4x^{3}}{3}+2x^{2}+\int -16\mathrm{d}x$$

Find the integral of -16 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{4x^{3}}{3}+2x^{2}-16x$$

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.

$$\frac{4}{3}\times 2^{3}+2\times 2^{2}-16\times 2-\left(\frac{4}{3}\left(-4\right)^{3}+2\left(-4\right)^{2}-16\left(-4\right)\right)$$

Simplify.

$$-24$$