Solve ∫ (from 0 to 2) of (8e^x-5) wrt x

$$\int _ { 0 } ^ { 2 } ( 8 e ^ { x } - 5 ) d x$$

Evaluate

$$8e^{2}-18\approx 41.112448791$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 8e^{x}-5\mathrm{d}x$$

Integrate the sum term by term.

$$\int 8e^{x}\mathrm{d}x+\int -5\mathrm{d}x$$

Factor out the constant in each of the terms.

$$8\int e^{x}\mathrm{d}x+\int -5\mathrm{d}x$$

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

$$8e^{x}+\int -5\mathrm{d}x$$

Find the integral of -5 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$8e^{x}-5x$$

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.

$$8e^{2}-5\times 2-\left(8e^{0}-5\times 0\right)$$

Simplify.

$$8e^{2}-18$$