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

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

Evaluate

$$\frac{21956}{105}\approx 209.104761905$$

Solution Steps

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

$$\int _{1}^{2}4\left(x^{3}\right)^{2}+16x^{3}x+16x^{2}\mathrm{d}x$$

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

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

To multiply powers of the same base, add their exponents. Add 3 and 1 to get 4.

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

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{6}\mathrm{d}x with \frac{x^{7}}{7}. Multiply 4 times \frac{x^{7}}{7}.

$$\frac{4x^{7}}{7}+16\int x^{4}\mathrm{d}x+16\int x^{2}\mathrm{d}x$$

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 16 times \frac{x^{5}}{5}.

$$\frac{4x^{7}}{7}+\frac{16x^{5}}{5}+16\int x^{2}\mathrm{d}x$$

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

$$\frac{4x^{7}}{7}+\frac{16x^{5}}{5}+\frac{16x^{3}}{3}$$

Simplify.

$$\frac{16x^{3}}{3}+\frac{16x^{5}}{5}+\frac{4x^{7}}{7}$$

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.

$$\frac{16}{3}\times 2^{3}+\frac{16}{5}\times 2^{5}+\frac{4}{7}\times 2^{7}-\left(\frac{16}{3}\times 1^{3}+\frac{16}{5}\times 1^{5}+\frac{4}{7}\times 1^{7}\right)$$

Simplify.

$$\frac{21956}{105}$$