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

$$\int_{ 1 }^{ 4 } { \left(-x \right) }^{ 2 } +6x-6 d x$$

Evaluate

$$48$$

Solution Steps

Calculate -x to the power of 2 and get x^{2}.

$$\int _{1}^{4}x^{2}+6x-6\mathrm{d}x$$

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

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}.

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

Find the integral of -6 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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.

$$\frac{4^{3}}{3}+3\times 4^{2}-6\times 4-\left(\frac{1^{3}}{3}+3\times 1^{2}-6\right)$$

Simplify.

$$48$$