Solve ∫ (from 0 to 6) of (x^2+12x-4) wrt x

$$\int _ { 0 } ^ { 6 } ( x ^ { 2 } + 12 x - 4 ) d x$$

Evaluate

$$264$$

Solution Steps

Evaluate the indefinite integral first.

$$\int x^{2}+12x-4\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{2}\mathrm{d}x+\int 12x\mathrm{d}x+\int -4\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{2}\mathrm{d}x+12\int x\mathrm{d}x+\int -4\mathrm{d}x$$

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}.

$$\frac{x^{3}}{3}+12\int x\mathrm{d}x+\int -4\mathrm{d}x$$

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

$$\frac{x^{3}}{3}+6x^{2}+\int -4\mathrm{d}x$$

Find the integral of -4 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{3}}{3}+6x^{2}-4x$$

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.

$$\frac{6^{3}}{3}+6\times 6^{2}-4\times 6-\left(\frac{0^{3}}{3}+6\times 0^{2}-4\times 0\right)$$

Simplify.

$$264$$