Solve ∫ (from 0 to 2 pi) of (3x+2y) wrt y

$$\int _ { 0 } ^ { 2 \pi } ( 3 x + 2 y ) d y$$

Evaluate

$$2\pi \left(3x+2\pi \right)$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

$$3xy+2\int y\mathrm{d}y$$

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}. Multiply 2 times \frac{y^{2}}{2}.

$$3xy+y^{2}$$

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.

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

Simplify.

$$2\pi \left(3x+2\pi \right)$$

Differentiate w.r.t. x

$$6\pi$$