Solve ∫ (from 2 to 9) of (sqrt{13^2}-(1/3)^3)^2=

$$\int _ { 2 } ^ { 9 } ( \sqrt { 13 ^ { 2 } } - ( \frac { 1 } { 3 } ) ^ { 3 } ) ^ { 2 } =$$

Evaluate

$$\frac{857500}{729}\approx 1176.268861454$$

Solution Steps

Calculate 13 to the power of 2 and get 169.

$$\int _{2}^{9}\left(\sqrt{169}-\left(\frac{1}{3}\right)^{3}\right)^{2}\mathrm{d}x$$

Calculate the square root of 169 and get 13.

$$\int _{2}^{9}\left(13-\left(\frac{1}{3}\right)^{3}\right)^{2}\mathrm{d}x$$

Calculate \frac{1}{3} to the power of 3 and get \frac{1}{27}.

$$\int _{2}^{9}\left(13-\frac{1}{27}\right)^{2}\mathrm{d}x$$

Subtract \frac{1}{27} from 13 to get \frac{350}{27}.

$$\int _{2}^{9}\left(\frac{350}{27}\right)^{2}\mathrm{d}x$$

Calculate \frac{350}{27} to the power of 2 and get \frac{122500}{729}.

$$\int _{2}^{9}\frac{122500}{729}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int \frac{122500}{729}\mathrm{d}x$$

Find the integral of \frac{122500}{729} using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{122500x}{729}$$

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.

$$\frac{122500}{729}\times 9-\frac{122500}{729}\times 2$$

Simplify.

$$\frac{857500}{729}$$