Solve ∫ (from 0 to 1) of int_{y^2}^1int_0^1-zz wrt ydzdx

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

Evaluate

$$\frac{y^{6}}{3}-\frac{y^{4}}{2}+\frac{1}{6}$$

Solution Steps

Evaluate the indefinite integral first.

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

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

$$\int _{y^{2}}^{1}\int _{0}^{1-z}z\mathrm{d}y\mathrm{d}zx$$

Simplify.

$$\left(\frac{1}{6}-\frac{y^{4}}{2}+\frac{y^{6}}{3}\right)x$$

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.

$$\left(\frac{1}{6}-\frac{1}{2}y^{4}+\frac{1}{3}y^{6}\right)\times 1-\left(\frac{1}{6}-\frac{1}{2}y^{4}+\frac{1}{3}y^{6}\right)\times 0$$

Simplify.

$$\frac{1}{6}-\frac{y^{4}}{2}+\frac{y^{6}}{3}$$