Solve ∫ (from 0 to 1) of 4x(1+x)^3 wrt x

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

Evaluate

$$\frac{49}{5}=9.8$$

Solution Steps

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

$$\int _{0}^{1}4x\left(1+3x+3x^{2}+x^{3}\right)\mathrm{d}x$$

Use the distributive property to multiply 4x by 1+3x+3x^{2}+x^{3}.

$$\int _{0}^{1}4x+12x^{2}+12x^{3}+4x^{4}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 4x+12x^{2}+12x^{3}+4x^{4}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 4x\mathrm{d}x+\int 12x^{2}\mathrm{d}x+\int 12x^{3}\mathrm{d}x+\int 4x^{4}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$4\int x\mathrm{d}x+12\int x^{2}\mathrm{d}x+12\int x^{3}\mathrm{d}x+4\int x^{4}\mathrm{d}x$$

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

$$2x^{2}+12\int x^{2}\mathrm{d}x+12\int x^{3}\mathrm{d}x+4\int x^{4}\mathrm{d}x$$

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}.

$$2x^{2}+4x^{3}+12\int x^{3}\mathrm{d}x+4\int x^{4}\mathrm{d}x$$

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

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

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{4}\mathrm{d}x with \frac{x^{5}}{5}. Multiply 4 times \frac{x^{5}}{5}.

$$2x^{2}+4x^{3}+3x^{4}+\frac{4x^{5}}{5}$$

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.

$$2\times 1^{2}+4\times 1^{3}+3\times 1^{4}+\frac{4}{5}\times 1^{5}-\left(2\times 0^{2}+4\times 0^{3}+3\times 0^{4}+\frac{4}{5}\times 0^{5}\right)$$

Simplify.

$$\frac{49}{5}$$