Solve ∫ (from 1 to 6) of 2x^2-16x+32 wrt x

$$\int _ { 1 } ^ { 6 } 2 x ^ { 2 } - 16 x + 32 d x$$

Evaluate

$$\frac{70}{3}\approx 23.333333333$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 2x^{2}-16x+32\mathrm{d}x$$

Integrate the sum term by term.

$$\int 2x^{2}\mathrm{d}x+\int -16x\mathrm{d}x+\int 32\mathrm{d}x$$

Factor out the constant in each of the terms.

$$2\int x^{2}\mathrm{d}x-16\int x\mathrm{d}x+\int 32\mathrm{d}x$$

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}.

$$\frac{2x^{3}}{3}-16\int x\mathrm{d}x+\int 32\mathrm{d}x$$

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 -16 times \frac{x^{2}}{2}.

$$\frac{2x^{3}}{3}-8x^{2}+\int 32\mathrm{d}x$$

Find the integral of 32 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{2x^{3}}{3}-8x^{2}+32x$$

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.

$$\frac{2}{3}\times 6^{3}-8\times 6^{2}+32\times 6-\left(\frac{2}{3}\times 1^{3}-8\times 1^{2}+32\times 1\right)$$

Simplify.

$$\frac{70}{3}$$