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

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

Evaluate

$$4$$

Solution Steps

To find the opposite of -3x^{2}+6x, find the opposite of each term.

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

Combine -6x^{2} and 3x^{2} to get -3x^{2}.

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

Combine 12x and -6x to get 6x.

$$\int _{0}^{2}-3x^{2}+6x\mathrm{d}x$$

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

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

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}.

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

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.

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

Simplify.

$$4$$