Solve ∫ (from - 1 to 0) of (-1/2x^2+1-2^x) wrt x=

$$\int _ { - 1 } ^ { 0 } ( - \frac { 1 } { 2 } x ^ { 2 } + 1 - 2 ^ { x } ) d x =$$

Evaluate

$$-\frac{1}{2\ln(2)}+\frac{5}{6}\approx 0.111985813$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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 -\frac{1}{2} times \frac{x^{3}}{3}.

$$-\frac{x^{3}}{6}+\int 1\mathrm{d}x-\int 2^{x}\mathrm{d}x$$

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

$$-\frac{x^{3}}{6}+x-\int 2^{x}\mathrm{d}x$$

Use \int x^{k}\mathrm{d}k=\frac{x^{k}}{\ln(x)} from the table of common integrals to obtain the result.

$$-\frac{x^{3}}{6}+x-\frac{2^{x}}{\ln(2)}$$

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.

$$-\frac{0^{3}}{6}+0-2^{0}\ln(2)^{-1}-\left(-\frac{\left(-1\right)^{3}}{6}-1-2^{-1}\ln(2)^{-1}\right)$$

Simplify.

$$-\frac{1}{2\ln(2)}+\frac{5}{6}$$