Solve ∫ (from - 3 to 1 / 2) of (6x^2+4x) wrt x

$$\int _ { - 3 } ^ { 1 / 2 } ( 6 x ^ { 2 } + 4 x ) d x$$

Evaluate

$$\frac{147}{4}=36.75$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 6x^{2}+4x\mathrm{d}x$$

Integrate the sum term by term.

$$\int 6x^{2}\mathrm{d}x+\int 4x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$6\int x^{2}\mathrm{d}x+4\int x\mathrm{d}x$$

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}.

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

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

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

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.

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

Simplify.

$$\frac{147}{4}$$