Solve ∫ (from 1 to 2) of 2y-y^2 wrt y

$$\int_{ 1 }^{ 2 } 2y- { y }^{ 2 } d y$$

Evaluate

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

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

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}. Multiply -1 times \frac{y^{3}}{3}.

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

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.

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

Simplify.

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