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

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

Evaluate

$$\frac{9}{2}=4.5$$

Solution Steps

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

$$\int _{1}^{4}x-\left(x^{2}-4x+4\right)\mathrm{d}x$$

To find the opposite of x^{2}-4x+4, find the opposite of each term.

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

Combine x and 4x to get 5x.

$$\int _{1}^{4}5x-x^{2}-4\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 5x-x^{2}-4\mathrm{d}x$$

Integrate the sum term by term.

$$\int 5x\mathrm{d}x+\int -x^{2}\mathrm{d}x+\int -4\mathrm{d}x$$

Factor out the constant in each of the terms.

$$5\int x\mathrm{d}x-\int x^{2}\mathrm{d}x+\int -4\mathrm{d}x$$

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

$$\frac{5x^{2}}{2}-\int x^{2}\mathrm{d}x+\int -4\mathrm{d}x$$

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 -1 times \frac{x^{3}}{3}.

$$\frac{5x^{2}}{2}-\frac{x^{3}}{3}+\int -4\mathrm{d}x$$

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

$$\frac{5x^{2}}{2}-\frac{x^{3}}{3}-4x$$

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.

$$\frac{5}{2}\times 4^{2}-\frac{4^{3}}{3}-4\times 4-\left(\frac{5}{2}\times 1^{2}-\frac{1^{3}}{3}-4\right)$$

Simplify.

$$\frac{9}{2}$$