Solve ∫ (from 2 to 4) of 2x^5+3x wrt x

$$\int _ { 2 } ^ { 4 } 2 x ^ { 5 } + 3 x d x$$

Evaluate

$$1362$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 2x^{5}+3x\mathrm{d}x$$

Integrate the sum term by term.

$$\int 2x^{5}\mathrm{d}x+\int 3x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$2\int x^{5}\mathrm{d}x+3\int x\mathrm{d}x$$

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 2 times \frac{x^{6}}{6}.

$$\frac{x^{6}}{3}+3\int x\mathrm{d}x$$

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

$$\frac{x^{6}}{3}+\frac{3x^{2}}{2}$$

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.

$$\frac{4^{6}}{3}+\frac{3}{2}\times 4^{2}-\left(\frac{2^{6}}{3}+\frac{3}{2}\times 2^{2}\right)$$

Simplify.

$$1362$$