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

$$\int _ { 0 } ^ { 1 } x ^ { 2 } ( 1 + 3 x ^ { 3 } ) d x$$

Evaluate

$$\frac{5}{6}\approx 0.833333333$$

Solution Steps

Use the distributive property to multiply x^{2} by 1+3x^{3}.

$$\int _{0}^{1}x^{2}+3x^{5}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int x^{2}+3x^{5}\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{2}\mathrm{d}x+\int 3x^{5}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{2}\mathrm{d}x+3\int x^{5}\mathrm{d}x$$

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

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

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}. Multiply 3 times \frac{x^{6}}{6}.

$$\frac{x^{3}}{3}+\frac{x^{6}}{2}$$

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.

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

Simplify.

$$\frac{5}{6}$$