Solve ∫ (from 0 to 1) of (-x)^3+(-x)^2=

$$\int _ { 0 } ^ { 1 } ( - x ) ^ { 3 } + ( - x ) ^ { 2 } =$$

Evaluate

$$\frac{1}{12}\approx 0.083333333$$

Solution Steps

Calculate -x to the power of 2 and get x^{2}.

$$\int _{0}^{1}\left(-x\right)^{3}+x^{2}\mathrm{d}x$$

Expand \left(-x\right)^{3}.

$$\int _{0}^{1}\left(-1\right)^{3}x^{3}+x^{2}\mathrm{d}x$$

Calculate -1 to the power of 3 and get -1.

$$\int _{0}^{1}-x^{3}+x^{2}\mathrm{d}x$$

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{3}\mathrm{d}x with \frac{x^{4}}{4}. Multiply -1 times \frac{x^{4}}{4}.

$$-\frac{x^{4}}{4}+\int x^{2}\mathrm{d}x$$

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}.

$$-\frac{x^{4}}{4}+\frac{x^{3}}{3}$$

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.

$$-\frac{1^{4}}{4}+\frac{1^{3}}{3}-\left(-\frac{0^{4}}{4}+\frac{0^{3}}{3}\right)$$

Simplify.

$$\frac{1}{12}$$