Solve ∫ _{-frac{sqrt{38}}{6}-1/2}^sqrt{33/6-1/2}(-3x^2+2-3x) wrt x

$$\int _ { - \frac { \sqrt { 38 } } { 6 } - \frac { 1 } { 2 } } ^ { \frac { \sqrt { 33 } } { 6 } - \frac { 1 } { 2 } } ( - 3 x ^ { 2 } + 2 - 3 x ) d x$$

Evaluate

$$\frac{11\sqrt{33}}{36}+\frac{61\sqrt{38}}{216}\approx 3.496159208$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

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

Find the integral of 2 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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}.

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

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.

$$-\left(\frac{1}{6}\times 33^{\frac{1}{2}}-\frac{1}{2}\right)^{3}+2\left(\frac{1}{6}\times 33^{\frac{1}{2}}-\frac{1}{2}\right)-\frac{3}{2}\left(\frac{1}{6}\times 33^{\frac{1}{2}}-\frac{1}{2}\right)^{2}-\left(-\left(-\frac{1}{6}\times 38^{\frac{1}{2}}-\frac{1}{2}\right)^{3}+2\left(-\frac{1}{6}\times 38^{\frac{1}{2}}-\frac{1}{2}\right)-\frac{3}{2}\left(-\frac{1}{6}\times 38^{\frac{1}{2}}-\frac{1}{2}\right)^{2}\right)$$

Simplify.

$$\frac{11\sqrt{33}}{36}+\frac{61\sqrt{38}}{216}$$