Solve ∫ (from 0 to 2) of x^2(4+x^3) wrt x

$$\int _ { 0 } ^ { 2 } x ^ { 2 } ( 4 + x ^ { 3 } ) d x$$

Evaluate

$$\frac{64}{3}\approx 21.333333333$$

Solution Steps

Use the distributive property to multiply x^{2} by 4+x^{3}.

$$\int _{0}^{2}4x^{2}+x^{5}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 4x^{2}+x^{5}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 4x^{2}\mathrm{d}x+\int x^{5}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$4\int x^{2}\mathrm{d}x+\int x^{5}\mathrm{d}x$$

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}. Multiply 4 times \frac{x^{3}}{3}.

$$\frac{4x^{3}}{3}+\int x^{5}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{5}\mathrm{d}x with \frac{x^{6}}{6}.

$$\frac{4x^{3}}{3}+\frac{x^{6}}{6}$$

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.

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

Simplify.

$$\frac{64}{3}$$