Solve ∫ (from 0 to frac { pi) of {2}}x^3+x^2 wrt x

$$\int _ { 0 } ^ { \frac { \pi } { 2 } } x ^ { 3 } + x ^ { 2 } d x$$

Evaluate

$$\frac{\pi ^{3}\left(3\pi +8\right)}{192}\approx 2.813945242$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{3}\mathrm{d}x with \frac{x^{4}}{4}.

$$\frac{x^{4}}{4}+\int x^{2}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}.

$$\frac{x^{4}}{4}+\frac{x^{3}}{3}$$

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.

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

Simplify.

$$\frac{\left(3\pi +8\right)\pi ^{3}}{192}$$