Solve ∫ (from 0 to 2) of x(2+x^5) wrt x

$$\int _ { 0 } ^ { 2 } x ( 2 + x ^ { 5 } ) d x$$

Evaluate

$$\frac{156}{7}\approx 22.285714286$$

Solution Steps

Use the distributive property to multiply x by 2+x^{5}.

$$\int _{0}^{2}2x+x^{6}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 2x+x^{6}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 2x\mathrm{d}x+\int x^{6}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$2\int x\mathrm{d}x+\int x^{6}\mathrm{d}x$$

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

$$x^{2}+\int x^{6}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{6}\mathrm{d}x with \frac{x^{7}}{7}.

$$x^{2}+\frac{x^{7}}{7}$$

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.

$$2^{2}+\frac{2^{7}}{7}-\left(0^{2}+\frac{0^{7}}{7}\right)$$

Simplify.

$$\frac{156}{7}$$