Solve ∫ (from - 2 to 5) of (4x-3)^2 wrt x

$$\int _ { - 2 } ^ { 5 } ( 4 x - 3 ) ^ { 2 } d x$$

Evaluate

$$\frac{1561}{3}\approx 520.333333333$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4x-3\right)^{2}.

$$\int _{-2}^{5}16x^{2}-24x+9\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 16x^{2}-24x+9\mathrm{d}x$$

Integrate the sum term by term.

$$\int 16x^{2}\mathrm{d}x+\int -24x\mathrm{d}x+\int 9\mathrm{d}x$$

Factor out the constant in each of the terms.

$$16\int x^{2}\mathrm{d}x-24\int x\mathrm{d}x+\int 9\mathrm{d}x$$

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 16 times \frac{x^{3}}{3}.

$$\frac{16x^{3}}{3}-24\int x\mathrm{d}x+\int 9\mathrm{d}x$$

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 -24 times \frac{x^{2}}{2}.

$$\frac{16x^{3}}{3}-12x^{2}+\int 9\mathrm{d}x$$

Find the integral of 9 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{16x^{3}}{3}-12x^{2}+9x$$

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.

$$\frac{16}{3}\times 5^{3}-12\times 5^{2}+9\times 5-\left(\frac{16}{3}\left(-2\right)^{3}-12\left(-2\right)^{2}+9\left(-2\right)\right)$$

Simplify.

$$\frac{1561}{3}$$