Solve ∫ (from 0 to 2) of (6x^2-3x+12) wrt x

$$\int _ { 0 } ^ { 2 } ( 6 x ^ { 2 } - 3 x + 12 ) d x$$

Evaluate

$$34$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 6x^{2}-3x+12\mathrm{d}x$$

Integrate the sum term by term.

$$\int 6x^{2}\mathrm{d}x+\int -3x\mathrm{d}x+\int 12\mathrm{d}x$$

Factor out the constant in each of the terms.

$$6\int x^{2}\mathrm{d}x-3\int x\mathrm{d}x+\int 12\mathrm{d}x$$

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}.

$$2x^{3}-3\int x\mathrm{d}x+\int 12\mathrm{d}x$$

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 -3 times \frac{x^{2}}{2}.

$$2x^{3}-\frac{3x^{2}}{2}+\int 12\mathrm{d}x$$

Find the integral of 12 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$2x^{3}-\frac{3x^{2}}{2}+12x$$

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.

$$2\times 2^{3}-\frac{3}{2}\times 2^{2}+12\times 2-\left(2\times 0^{3}-\frac{3}{2}\times 0^{2}+12\times 0\right)$$

Simplify.

$$34$$