Solve ∫ (from 2 to 3) of (2x3+6x) wrt x=

$$\int _ { 2 } ^ { 3 } ( 2 x 3 + 6 x ) d x =$$

Evaluate

$$2x_{3}+15$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 2x_{3}+6x\mathrm{d}x$$

Integrate the sum term by term.

$$\int 2x_{3}\mathrm{d}x+\int 6x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$2\int x_{3}\mathrm{d}x+6\int x\mathrm{d}x$$

Find the integral of x_{3} using the table of common integrals rule \int a\mathrm{d}x=ax.

$$2x_{3}x+6\int x\mathrm{d}x$$

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 6 times \frac{x^{2}}{2}.

$$2x_{3}x+3x^{2}$$

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.

$$2x_{3}\times 3+3\times 3^{2}-\left(2x_{3}\times 2+3\times 2^{2}\right)$$

Simplify.

$$2x_{3}+15$$

Differentiate w.r.t. x_3

$$2$$