Solve ∫ (from 1 to 6) of (5x^2+6x+10) wrt x

$$\int _ { 1 } ^ { 6 } ( 5 x ^ { 2 } + 6 x + 10 ) d x$$

Evaluate

$$\frac{1540}{3}\approx 513.333333333$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 5x^{2}+6x+10\mathrm{d}x$$

Integrate the sum term by term.

$$\int 5x^{2}\mathrm{d}x+\int 6x\mathrm{d}x+\int 10\mathrm{d}x$$

Factor out the constant in each of the terms.

$$5\int x^{2}\mathrm{d}x+6\int x\mathrm{d}x+\int 10\mathrm{d}x$$

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

$$\frac{5x^{3}}{3}+6\int x\mathrm{d}x+\int 10\mathrm{d}x$$

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

$$\frac{5x^{3}}{3}+3x^{2}+\int 10\mathrm{d}x$$

Find the integral of 10 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{5x^{3}}{3}+3x^{2}+10x$$

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.

$$\frac{5}{3}\times 6^{3}+3\times 6^{2}+10\times 6-\left(\frac{5}{3}\times 1^{3}+3\times 1^{2}+10\times 1\right)$$

Simplify.

$$\frac{1540}{3}$$