Solve ∫ (from 0 to frac { 1) of {2}}(4x+1)^3 wrt x=

$$\int _ { 0 } ^ { \frac { 1 } { 2 } } ( 4 x + 1 ) ^ { 3 } d x =$$

Evaluate

$$5$$

Solution Steps

Use binomial theorem \left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} to expand \left(4x+1\right)^{3}.

$$\int _{0}^{\frac{1}{2}}64x^{3}+48x^{2}+12x+1\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 64x^{3}+48x^{2}+12x+1\mathrm{d}x$$

Integrate the sum term by term.

$$\int 64x^{3}\mathrm{d}x+\int 48x^{2}\mathrm{d}x+\int 12x\mathrm{d}x+\int 1\mathrm{d}x$$

Factor out the constant in each of the terms.

$$64\int x^{3}\mathrm{d}x+48\int x^{2}\mathrm{d}x+12\int x\mathrm{d}x+\int 1\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{3}\mathrm{d}x with \frac{x^{4}}{4}. Multiply 64 times \frac{x^{4}}{4}.

$$16x^{4}+48\int x^{2}\mathrm{d}x+12\int x\mathrm{d}x+\int 1\mathrm{d}x$$

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

$$16x^{4}+16x^{3}+12\int x\mathrm{d}x+\int 1\mathrm{d}x$$

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

$$16x^{4}+16x^{3}+6x^{2}+\int 1\mathrm{d}x$$

Find the integral of 1 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$16x^{4}+16x^{3}+6x^{2}+x$$

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.

$$16\times \left(\frac{1}{2}\right)^{4}+16\times \left(\frac{1}{2}\right)^{3}+6\times \left(\frac{1}{2}\right)^{2}+\frac{1}{2}-\left(16\times 0^{4}+16\times 0^{3}+6\times 0^{2}+0\right)$$

Simplify.

$$5$$