Solve ∫ (from 2 to 4) of (x+8)^2 wrt x=

$$\int _ { 2 } ^ { 4 } ( x + 8 ) ^ { 2 } d x =$$

Evaluate

$$\frac{728}{3}\approx 242.666666667$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+8\right)^{2}.

$$\int _{2}^{4}x^{2}+16x+64\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int x^{2}+16x+64\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{2}\mathrm{d}x+\int 16x\mathrm{d}x+\int 64\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{2}\mathrm{d}x+16\int x\mathrm{d}x+\int 64\mathrm{d}x$$

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}.

$$\frac{x^{3}}{3}+16\int x\mathrm{d}x+\int 64\mathrm{d}x$$

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

$$\frac{x^{3}}{3}+8x^{2}+\int 64\mathrm{d}x$$

Find the integral of 64 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{3}}{3}+8x^{2}+64x$$

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.

$$\frac{4^{3}}{3}+8\times 4^{2}+64\times 4-\left(\frac{2^{3}}{3}+8\times 2^{2}+64\times 2\right)$$

Simplify.

$$\frac{728}{3}$$