Solve ∫ (from 1 to 2) of 1/3+y/6 wrt y

$$\int _ { 1 } ^ { 2 } \frac { 1 } { 3 } + \frac { y } { 6 } d y$$

Evaluate

$$\frac{7}{12}\approx 0.583333333$$

Solution Steps

Evaluate the indefinite integral first.

$$\int \frac{1}{3}+\frac{y}{6}\mathrm{d}y$$

Integrate the sum term by term.

$$\int \frac{1}{3}\mathrm{d}y+\int \frac{y}{6}\mathrm{d}y$$

Factor out the constant in each of the terms.

$$\int \frac{1}{3}\mathrm{d}y+\frac{\int y\mathrm{d}y}{6}$$

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

$$\frac{y}{3}+\frac{\int y\mathrm{d}y}{6}$$

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 \frac{1}{6} times \frac{y^{2}}{2}.

$$\frac{y}{3}+\frac{y^{2}}{12}$$

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.

$$\frac{1}{3}\times 2+\frac{2^{2}}{12}-\left(\frac{1}{3}\times 1+\frac{1^{2}}{12}\right)$$

Simplify.

$$\frac{7}{12}$$