Solve ∫ (from - 4 to 2) of (-24x^2+96) wrt x

$$\int _ { - 4 } ^ { 2 } ( - 24 x ^ { 2 } + 96 ) d x$$

Evaluate

$$0$$

Solution Steps

Evaluate the indefinite integral first.

$$\int -24x^{2}+96\mathrm{d}x$$

Integrate the sum term by term.

$$\int -24x^{2}\mathrm{d}x+\int 96\mathrm{d}x$$

Factor out the constant in each of the terms.

$$-24\int x^{2}\mathrm{d}x+\int 96\mathrm{d}x$$

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 -24 times \frac{x^{3}}{3}.

$$-8x^{3}+\int 96\mathrm{d}x$$

Find the integral of 96 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$-8x^{3}+96x$$

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.

$$-8\times 2^{3}+96\times 2-\left(-8\left(-4\right)^{3}+96\left(-4\right)\right)$$

Simplify.

$$0$$