Solve ∫ (from 0 to 1) of (y^2-2y) wrt y

$$\int _ { 0 } ^ { 1 } ( y ^ { 2 } - 2 y ) d y$$

Evaluate

$$-\frac{2}{3}\approx -0.666666667$$

Solution Steps

Evaluate the indefinite integral first.

$$\int y^{2}-2y\mathrm{d}y$$

Integrate the sum term by term.

$$\int y^{2}\mathrm{d}y+\int -2y\mathrm{d}y$$

Factor out the constant in each of the terms.

$$\int y^{2}\mathrm{d}y-2\int y\mathrm{d}y$$

Since \int y^{k}\mathrm{d}y=\frac{y^{k+1}}{k+1} for k\neq -1, replace \int y^{2}\mathrm{d}y with \frac{y^{3}}{3}.

$$\frac{y^{3}}{3}-2\int y\mathrm{d}y$$

Since \int y^{k}\mathrm{d}y=\frac{y^{k+1}}{k+1} for k\neq -1, replace \int y\mathrm{d}y with \frac{y^{2}}{2}. Multiply -2 times \frac{y^{2}}{2}.

$$\frac{y^{3}}{3}-y^{2}$$

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.

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

Simplify.

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