Solve ∫ (from 2 to 6) of 2x^2-3sqrt{x wrt x

$$\int_{ 2 }^{ 6 } 2 { x }^{ 2 } -3 \sqrt{ x } d x$$

Evaluate

$$4\sqrt{2}+\frac{416}{3}-12\sqrt{6}\approx 114.929644003$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

Rewrite \sqrt{x} as x^{\frac{1}{2}}. Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{\frac{1}{2}}\mathrm{d}x with \frac{x^{\frac{3}{2}}}{\frac{3}{2}}. Simplify. Multiply -3 times \frac{2x^{\frac{3}{2}}}{3}.

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

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.

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

Simplify.

$$\frac{416}{3}-12\sqrt{6}+4\sqrt{2}$$