Solve ∫ (from 0 to 2) of int_{x^2}^2x1 wrt ydx

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

Evaluate

$$\frac{4}{3}\approx 1.333333333$$

Solution Steps

Evaluate the indefinite integral first.

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

Simplify.

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

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.

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

Simplify.

$$\frac{4}{3}$$