Solve ∫ (from - 2 to 0) of (16x-e^x) wrt x

$$\int _ { - 2 } ^ { 0 } ( 16 x - e ^ { x } ) d x$$

Evaluate

$$\frac{1}{e^{2}}-33\approx -32.864664717$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 16x-e^{x}\mathrm{d}x$$

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

$$16\int x\mathrm{d}x-\int e^{x}\mathrm{d}x$$

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 16 times \frac{x^{2}}{2}.

$$8x^{2}-\int e^{x}\mathrm{d}x$$

Use \int e^{x}\mathrm{d}x=e^{x} from the table of common integrals to obtain the result.

$$8x^{2}-e^{x}$$

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.

$$8\times 0^{2}-e^{0}-\left(8\left(-2\right)^{2}-e^{-2}\right)$$

Simplify.

$$-33+\frac{1}{e^{2}}$$