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

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

Evaluate

$$\frac{65}{2}=32.5$$

Solution Steps

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

$$\int _{0}^{1}4x\left(\left(x^{2}\right)^{3}+6\left(x^{2}\right)^{2}+12x^{2}+8\right)\mathrm{d}x$$

To raise a power to another power, multiply the exponents. Multiply 2 and 3 to get 6.

$$\int _{0}^{1}4x\left(x^{6}+6\left(x^{2}\right)^{2}+12x^{2}+8\right)\mathrm{d}x$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$\int _{0}^{1}4x\left(x^{6}+6x^{4}+12x^{2}+8\right)\mathrm{d}x$$

Use the distributive property to multiply 4x by x^{6}+6x^{4}+12x^{2}+8.

$$\int _{0}^{1}4x^{7}+24x^{5}+48x^{3}+32x\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 4x^{7}+24x^{5}+48x^{3}+32x\mathrm{d}x$$

Integrate the sum term by term.

$$\int 4x^{7}\mathrm{d}x+\int 24x^{5}\mathrm{d}x+\int 48x^{3}\mathrm{d}x+\int 32x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$4\int x^{7}\mathrm{d}x+24\int x^{5}\mathrm{d}x+48\int x^{3}\mathrm{d}x+32\int x\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{7}\mathrm{d}x with \frac{x^{8}}{8}. Multiply 4 times \frac{x^{8}}{8}.

$$\frac{x^{8}}{2}+24\int x^{5}\mathrm{d}x+48\int x^{3}\mathrm{d}x+32\int x\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{5}\mathrm{d}x with \frac{x^{6}}{6}. Multiply 24 times \frac{x^{6}}{6}.

$$\frac{x^{8}}{2}+4x^{6}+48\int x^{3}\mathrm{d}x+32\int x\mathrm{d}x$$

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

$$\frac{x^{8}}{2}+4x^{6}+12x^{4}+32\int x\mathrm{d}x$$

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

$$\frac{x^{8}}{2}+4x^{6}+12x^{4}+16x^{2}$$

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.

$$16\times 1^{2}+12\times 1^{4}+4\times 1^{6}+\frac{1^{8}}{2}-\left(16\times 0^{2}+12\times 0^{4}+4\times 0^{6}+\frac{0^{8}}{2}\right)$$

Simplify.

$$\frac{65}{2}$$