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

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}.

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

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}. Multiply -1 times \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.

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