Solve ∫ (from 0 to frac { 1) of {2}}(2y-3y^2-y^2) wrt y

$$\int _ { 0 } ^ { \frac { 1 } { 2 } } ( 2 y - 3 y ^ { 2 } - y ^ { 2 } ) d y$$

Evaluate

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

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

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

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

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}.

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

Simplify.

$$y^{2}-\frac{4y^{3}}{3}$$

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.

$$\left(\frac{1}{2}\right)^{2}-\frac{4}{3}\times \left(\frac{1}{2}\right)^{3}-\left(0^{2}-\frac{4}{3}\times 0^{3}\right)$$

Simplify.

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