Solve ∫ (from 0 to 2) of x^2+1/3x^4 wrt x

$$\int _ { 0 } ^ { 2 } x ^ { 2 } + \frac { 1 } { 3 } x ^ { 4 } d x$$

Evaluate

$$\frac{24}{5}=4.8$$

Solution Steps

Evaluate the indefinite integral first.

$$\int x^{2}+\frac{x^{4}}{3}\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{2}\mathrm{d}x+\int \frac{x^{4}}{3}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{2}\mathrm{d}x+\frac{\int x^{4}\mathrm{d}x}{3}$$

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}.

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

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{4}\mathrm{d}x with \frac{x^{5}}{5}. Multiply \frac{1}{3} times \frac{x^{5}}{5}.

$$\frac{x^{3}}{3}+\frac{x^{5}}{15}$$

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.

$$\frac{2^{3}}{3}+\frac{2^{5}}{15}-\left(\frac{0^{3}}{3}+\frac{0^{5}}{15}\right)$$

Simplify.

$$\frac{24}{5}$$