Solve ∫ (from 0 to 2) of 1/16y^3+3/16y^3 wrt y

$$\int _ { 0 } ^ { 2 } \frac { 1 } { 16 } y ^ { 3 } + \frac { 3 } { 16 } y ^ { 3 } d y$$

Evaluate

$$1$$

Solution Steps

Evaluate the indefinite integral first.

$$\int \frac{y^{3}+3y^{3}}{16}\mathrm{d}y$$

Integrate the sum term by term.

$$\int \frac{y^{3}}{16}\mathrm{d}y+\int \frac{3y^{3}}{16}\mathrm{d}y$$

Factor out the constant in each of the terms.

$$\frac{\int y^{3}\mathrm{d}y+3\int y^{3}\mathrm{d}y}{16}$$

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}{16} times \frac{y^{4}}{4}.

$$\frac{y^{4}}{64}+\frac{3\int y^{3}\mathrm{d}y}{16}$$

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{3}{16} times \frac{y^{4}}{4}.

$$\frac{y^{4}+3y^{4}}{64}$$

Simplify.

$$\frac{y^{4}}{16}$$

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.

$$\frac{2^{4}}{16}-\frac{0^{4}}{16}$$

Simplify.

$$1$$