Solve ∫ (from 1 to 3) of (4x^3+5)^2

$$\int _ { 1 } ^ { 3 } ( 4 x ^ { 3 } + 5 ) ^ { 2 }$$

Evaluate

$$\frac{40926}{7}\approx 5846.571428571$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4x^{3}+5\right)^{2}.

$$\int _{1}^{3}16\left(x^{3}\right)^{2}+40x^{3}+25\mathrm{d}x$$

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

$$\int _{1}^{3}16x^{6}+40x^{3}+25\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 16x^{6}+40x^{3}+25\mathrm{d}x$$

Integrate the sum term by term.

$$\int 16x^{6}\mathrm{d}x+\int 40x^{3}\mathrm{d}x+\int 25\mathrm{d}x$$

Factor out the constant in each of the terms.

$$16\int x^{6}\mathrm{d}x+40\int x^{3}\mathrm{d}x+\int 25\mathrm{d}x$$

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 16 times \frac{x^{7}}{7}.

$$\frac{16x^{7}}{7}+40\int x^{3}\mathrm{d}x+\int 25\mathrm{d}x$$

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

$$\frac{16x^{7}}{7}+10x^{4}+\int 25\mathrm{d}x$$

Find the integral of 25 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{16x^{7}}{7}+10x^{4}+25x$$

Simplify.

$$25x+10x^{4}+\frac{16x^{7}}{7}$$

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.

$$25\times 3+10\times 3^{4}+\frac{16}{7}\times 3^{7}-\left(25\times 1+10\times 1^{4}+\frac{16}{7}\times 1^{7}\right)$$

Simplify.

$$\frac{40926}{7}$$