Solve ∫ (from 3 to 6) of 2x^2-4x+3 wrt x

$$\int _ { 3 } ^ { 6 } 2 x ^ { 2 } - 4 x + 3 d x$$

Evaluate

$$81$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 2x^{2}-4x+3\mathrm{d}x$$

Integrate the sum term by term.

$$\int 2x^{2}\mathrm{d}x+\int -4x\mathrm{d}x+\int 3\mathrm{d}x$$

Factor out the constant in each of the terms.

$$2\int x^{2}\mathrm{d}x-4\int x\mathrm{d}x+\int 3\mathrm{d}x$$

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}.

$$\frac{2x^{3}}{3}-4\int x\mathrm{d}x+\int 3\mathrm{d}x$$

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}.

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

Find the integral of 3 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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.

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

Simplify.

$$81$$