Solve ∫ (from 1 to 4) of (2x-2)^2 wrt x=

$$\int _ { 1 } ^ { 4 } ( 2 x - 2 ) ^ { 2 } d x =$$

Evaluate

$$36$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-2\right)^{2}.

$$\int _{1}^{4}4x^{2}-8x+4\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 4x^{2}-8x+4\mathrm{d}x$$

Integrate the sum term by term.

$$\int 4x^{2}\mathrm{d}x+\int -8x\mathrm{d}x+\int 4\mathrm{d}x$$

Factor out the constant in each of the terms.

$$4\int x^{2}\mathrm{d}x-8\int x\mathrm{d}x+\int 4\mathrm{d}x$$

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}. Multiply 4 times \frac{x^{3}}{3}.

$$\frac{4x^{3}}{3}-8\int x\mathrm{d}x+\int 4\mathrm{d}x$$

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 -8 times \frac{x^{2}}{2}.

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

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

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

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.

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

Simplify.

$$36$$