Solve ∫ (from 0 to 1) of (x^2+y^2+x) wrt y

$$\int_{ 0 }^{ 1 } ( { x }^{ 2 } + { y }^{ 2 } +x) d y$$

Evaluate

$$x^{2}+x+\frac{1}{3}$$

Solution Steps

Evaluate the indefinite integral first.

$$\int x^{2}+y^{2}+x\mathrm{d}y$$

Integrate the sum term by term.

$$\int x^{2}\mathrm{d}y+\int y^{2}\mathrm{d}y+\int x\mathrm{d}y$$

Find the integral of x^{2} using the table of common integrals rule \int a\mathrm{d}y=ay.

$$x^{2}y+\int y^{2}\mathrm{d}y+\int x\mathrm{d}y$$

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}.

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

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

$$x^{2}y+\frac{y^{3}}{3}+xy$$

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.

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

Simplify.

$$x^{2}+\frac{1}{3}+x$$

Differentiate w.r.t. x

$$2x+1$$