Solve ∫ (from 0 to 1) of 1/3-1/3y^3 wrt y

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

Evaluate

$$\frac{1}{4}=0.25$$

Solution Steps

Evaluate the indefinite integral first.

$$\int \frac{1-y^{3}}{3}\mathrm{d}y$$

Integrate the sum term by term.

$$\int \frac{1}{3}\mathrm{d}y+\int -\frac{y^{3}}{3}\mathrm{d}y$$

Factor out the constant in each of the terms.

$$\int \frac{1}{3}\mathrm{d}y-\frac{\int y^{3}\mathrm{d}y}{3}$$

Find the integral of \frac{1}{3} using the table of common integrals rule \int a\mathrm{d}y=ay.

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

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

$$\frac{y}{3}-\frac{y^{4}}{12}$$

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.

$$\frac{1}{3}\times 1-\frac{1^{4}}{12}-\left(\frac{1}{3}\times 0-\frac{0^{4}}{12}\right)$$

Simplify.

$$\frac{1}{4}$$