Solve ∫ (from 1 to 4) of left(x+4right)^2 wrt x

$$\int_{ 1 }^{ 4 } { \left(x+4 \right) }^{ 2 } d x$$

Evaluate

$$129$$

Solution Steps

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

$$\int _{1}^{4}x^{2}+8x+16\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int x^{2}+8x+16\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{2}\mathrm{d}x+\int 8x\mathrm{d}x+\int 16\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{2}\mathrm{d}x+8\int x\mathrm{d}x+\int 16\mathrm{d}x$$

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}.

$$\frac{x^{3}}{3}+8\int x\mathrm{d}x+\int 16\mathrm{d}x$$

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}.

$$\frac{x^{3}}{3}+4x^{2}+\int 16\mathrm{d}x$$

Find the integral of 16 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{3}}{3}+4x^{2}+16x$$

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.

$$\frac{4^{3}}{3}+4\times 4^{2}+16\times 4-\left(\frac{1^{3}}{3}+4\times 1^{2}+16\times 1\right)$$

Simplify.

$$129$$