Solve ∫ (from 0 to 2 pi) of int_0^2-x1 wrt ydθ

$$\int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { 2 - x } 1 d y d \theta$$

Evaluate

$$4\pi -2\pi x$$

Solution Steps

Evaluate the indefinite integral first.

$$\int \int _{0}^{2-x}1\mathrm{d}y\mathrm{d}\theta$$

Find the integral of \int _{0}^{2-x}1\mathrm{d}y using the table of common integrals rule \int a\mathrm{d}\theta =a\theta .

$$\int _{0}^{2-x}1\mathrm{d}y\theta$$

Simplify.

$$\left(2-x\right)\theta$$

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.

$$\left(2-x\right)\times 2\pi -\left(2-x\right)\times 0$$

Simplify.

$$4\pi -2\pi x$$

Differentiate w.r.t. x

$$-2\pi$$