Solve ∫ (from 2 to 3) of 6x^2-4x+7-1/x^2 wrt x=

$$\int _ { 2 } ^ { 3 } 6 x ^ { 2 } - 4 x + 7 - \frac { 1 } { x ^ { 2 } } d x =$$

Evaluate

$$\frac{209}{6}\approx 34.833333333$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 6x^{2}-4x+7-\frac{1}{x^{2}}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 6x^{2}\mathrm{d}x+\int -4x\mathrm{d}x+\int 7\mathrm{d}x+\int -\frac{1}{x^{2}}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$6\int x^{2}\mathrm{d}x-4\int x\mathrm{d}x+\int 7\mathrm{d}x-\int \frac{1}{x^{2}}\mathrm{d}x$$

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

$$2x^{3}-4\int x\mathrm{d}x+\int 7\mathrm{d}x-\int \frac{1}{x^{2}}\mathrm{d}x$$

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}.

$$2x^{3}-2x^{2}+\int 7\mathrm{d}x-\int \frac{1}{x^{2}}\mathrm{d}x$$

Find the integral of 7 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$2x^{3}-2x^{2}+7x-\int \frac{1}{x^{2}}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int \frac{1}{x^{2}}\mathrm{d}x with -\frac{1}{x}. Multiply -1 times -\frac{1}{x}.

$$2x^{3}-2x^{2}+7x+\frac{1}{x}$$

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.

$$2\times 3^{3}-2\times 3^{2}+7\times 3+3^{-1}-\left(2\times 2^{3}-2\times 2^{2}+7\times 2+2^{-1}\right)$$

Simplify.

$$\frac{209}{6}$$