Solve ∫ (from - 2 to 1) of 2x(3x+1) wrt x=

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

Evaluate

$$15$$

Solution Steps

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

$$\int _{-2}^{1}6x^{2}+2x\mathrm{d}x$$

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}. Multiply 6 times \frac{x^{3}}{3}.

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

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

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

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.

$$2\times 1^{3}+1^{2}-\left(2\left(-2\right)^{3}+\left(-2\right)^{2}\right)$$

Simplify.

$$15$$