Solve ∫ (from 0 to 1) of 1+y wrt y

$$\int _ { 0 } ^ { 1 } 1 + y d y$$

Evaluate

$$\frac{3}{2}=1.5$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 1+y\mathrm{d}y$$

Integrate the sum term by term.

$$\int 1\mathrm{d}y+\int y\mathrm{d}y$$

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

$$y+\int y\mathrm{d}y$$

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}.

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

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.

$$1+\frac{1^{2}}{2}-\left(0+\frac{0^{2}}{2}\right)$$

Simplify.

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