Solve ∫ (from - 1 to 2) of (2x-5)^2 wrt x=

$$\int _ { - 1 } ^ { 2 } ( 2 x - 5 ) ^ { 2 } d x =$$

Evaluate

$$57$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-5\right)^{2}.

$$\int _{-1}^{2}4x^{2}-20x+25\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 4x^{2}-20x+25\mathrm{d}x$$

Integrate the sum term by term.

$$\int 4x^{2}\mathrm{d}x+\int -20x\mathrm{d}x+\int 25\mathrm{d}x$$

Factor out the constant in each of the terms.

$$4\int x^{2}\mathrm{d}x-20\int x\mathrm{d}x+\int 25\mathrm{d}x$$

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}.

$$\frac{4x^{3}}{3}-20\int x\mathrm{d}x+\int 25\mathrm{d}x$$

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 -20 times \frac{x^{2}}{2}.

$$\frac{4x^{3}}{3}-10x^{2}+\int 25\mathrm{d}x$$

Find the integral of 25 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{4x^{3}}{3}-10x^{2}+25x$$

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.

$$\frac{4}{3}\times 2^{3}-10\times 2^{2}+25\times 2-\left(\frac{4}{3}\left(-1\right)^{3}-10\left(-1\right)^{2}+25\left(-1\right)\right)$$

Simplify.

$$57$$