Solve ∫ (from 0 to ln 2) of e^x-1 wrt x=

$$\int _ { 0 } ^ { \ln 2 } e ^ { x } - 1 d x =$$

Evaluate

$$1-\ln(2)\approx 0.306852819$$

Solution Steps

Evaluate the indefinite integral first.

$$\int e^{x}-1\mathrm{d}x$$

Integrate the sum term by term.

$$\int e^{x}\mathrm{d}x+\int -1\mathrm{d}x$$

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

$$e^{x}+\int -1\mathrm{d}x$$

Find the integral of -1 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$e^{x}-x$$

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.

$$e^{\ln(2)\ln(e)^{-1}}-\ln(2)\ln(e)^{-1}-\left(e^{0}-0\right)$$

Simplify.

$$1-\ln(2)$$