Solve ∫ (from 0 to 1) of (1+3x)^3*2 wrt x

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

Evaluate

$$\frac{85}{2}=42.5$$

Solution Steps

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

$$\int _{0}^{1}\left(1+9x+27x^{2}+27x^{3}\right)\times 2\mathrm{d}x$$

Use the distributive property to multiply 1+9x+27x^{2}+27x^{3} by 2.

$$\int _{0}^{1}2+18x+54x^{2}+54x^{3}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 2+18x+54x^{2}+54x^{3}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 2\mathrm{d}x+\int 18x\mathrm{d}x+\int 54x^{2}\mathrm{d}x+\int 54x^{3}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int 2\mathrm{d}x+18\int x\mathrm{d}x+54\int x^{2}\mathrm{d}x+54\int x^{3}\mathrm{d}x$$

Find the integral of 2 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$2x+18\int x\mathrm{d}x+54\int x^{2}\mathrm{d}x+54\int x^{3}\mathrm{d}x$$

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

$$2x+9x^{2}+54\int x^{2}\mathrm{d}x+54\int x^{3}\mathrm{d}x$$

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

$$2x+9x^{2}+18x^{3}+54\int x^{3}\mathrm{d}x$$

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

$$2x+9x^{2}+18x^{3}+\frac{27x^{4}}{2}$$

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.

$$2\times 1+9\times 1^{2}+18\times 1^{3}+\frac{27}{2}\times 1^{4}-\left(2\times 0+9\times 0^{2}+18\times 0^{3}+\frac{27}{2}\times 0^{4}\right)$$

Simplify.

$$\frac{85}{2}$$