Solve ∫ (from 0 to 1) of (e^x-(-x+1)) wrt x

$$\int_{ 0 }^{ 1 } ( { e }^{ x } -(-x+1)) d x$$

Evaluate

$$e = \frac{3}{2} = 1\frac{1}{2} \approx 1.218281828$$

Solution Steps

To find the opposite of -x+1, find the opposite of each term.

$$\int _{0}^{1}e^{x}-\left(-x\right)-1\mathrm{d}x$$

Multiply -1 and -1 to get 1.

$$\int _{0}^{1}e^{x}+x-1\mathrm{d}x$$

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

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

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

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}.

$$e^{x}+\frac{x^{2}}{2}+\int -1\mathrm{d}x$$

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

$$e^{x}+\frac{x^{2}}{2}-x$$

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.

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

Simplify.

$$e-\frac{3}{2}$$