Solve ∫ (from 0 to 52) of x^2+x^3 wrt x

$$\int_{ 0 }^{ 52 } { x }^{ 2 } + { x }^{ 3 } d x$$

Evaluate

$$\frac{5624320}{3}\approx 1874773.333333333$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

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}.

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

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}.

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

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.

$$\frac{52^{3}}{3}+\frac{52^{4}}{4}-\left(\frac{0^{3}}{3}+\frac{0^{4}}{4}\right)$$

Simplify.

$$\frac{5624320}{3}$$