Solve ∫ (from 0 to 1) of (2x+1)^3

$$\int _ { 0 } ^ { 1 } ( 2 x + 1 ) ^ { 3 }$$

Evaluate

$$10$$

Solution Steps

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

$$\int _{0}^{1}8x^{3}+12x^{2}+6x+1\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 8x^{3}+12x^{2}+6x+1\mathrm{d}x$$

Integrate the sum term by term.

$$\int 8x^{3}\mathrm{d}x+\int 12x^{2}\mathrm{d}x+\int 6x\mathrm{d}x+\int 1\mathrm{d}x$$

Factor out the constant in each of the terms.

$$8\int x^{3}\mathrm{d}x+12\int x^{2}\mathrm{d}x+6\int x\mathrm{d}x+\int 1\mathrm{d}x$$

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 8 times \frac{x^{4}}{4}.

$$2x^{4}+12\int x^{2}\mathrm{d}x+6\int x\mathrm{d}x+\int 1\mathrm{d}x$$

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

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

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

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

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

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

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.

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

Simplify.

$$10$$