Solve ∫ (from -0.5 to 1) of (x+1-2x^2) wrt x

$$\int_{ -0.5 }^{ 1 } (x+1-2 { x }^{ 2 } ) d x$$

Evaluate

$$1.125$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

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

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

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

$$\frac{x^{2}}{2}+x-\frac{2x^{3}}{3}$$

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.

$$\frac{1^{2}}{2}+1-\frac{2}{3}\times 1^{3}-\left(\frac{\left(-0.5\right)^{2}}{2}-0.5-\frac{2}{3}\left(-0.5\right)^{3}\right)$$

Simplify.

$$\frac{9}{8}$$