Solve ∫ (from 3 to 5) of 4x+16x^3

$$\int _ { 3 } ^ { 5 } 4 x + 16 x ^ { 3 }$$

Evaluate

$$2208$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 4x+16x^{3}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 4x\mathrm{d}x+\int 16x^{3}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$4\int x\mathrm{d}x+16\int x^{3}\mathrm{d}x$$

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

$$2x^{2}+16\int x^{3}\mathrm{d}x$$

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

$$2x^{2}+4x^{4}$$

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.

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

Simplify.

$$2208$$