Solve ∫ (from 0 to 6) of 2x^2+3x+6 wrt x

$$\int _ { 0 } ^ { 6 } 2 x ^ { 2 } + 3 x + 6 d x$$

Evaluate

$$234$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 2x^{2}+3x+6\mathrm{d}x$$

Integrate the sum term by term.

$$\int 2x^{2}\mathrm{d}x+\int 3x\mathrm{d}x+\int 6\mathrm{d}x$$

Factor out the constant in each of the terms.

$$2\int x^{2}\mathrm{d}x+3\int x\mathrm{d}x+\int 6\mathrm{d}x$$

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

$$\frac{2x^{3}}{3}+3\int x\mathrm{d}x+\int 6\mathrm{d}x$$

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}.

$$\frac{2x^{3}}{3}+\frac{3x^{2}}{2}+\int 6\mathrm{d}x$$

Find the integral of 6 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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.

$$\frac{2}{3}\times 6^{3}+\frac{3}{2}\times 6^{2}+6\times 6-\left(\frac{2}{3}\times 0^{3}+\frac{3}{2}\times 0^{2}+6\times 0\right)$$

Simplify.

$$234$$