Evaluating the Definite Integral

We are asked to evaluate the definite integral:

$$\int_{0}^{4} \left(\sqrt{25-x^2} - \frac{25}{2\sqrt{25-x^2}}\right) dx$$

Let's split the integral into two parts:

$$\int_{0}^{4} \sqrt{25-x^2} dx - \int_{0}^{4} \frac{25}{2\sqrt{25-x^2}} dx$$

Let $$I_1 = \int_{0}^{4} \sqrt{25-x^2} dx$$ and $$I_2 = \int_{0}^{4} \frac{25}{2\sqrt{25-x^2}} dx$$

So, we need to evaluate $$I_1 - I_2$$

Evaluating $$I_1$$

$$I_1 = \int_{0}^{4} \sqrt{25-x^2} dx$$

Let $$x = 5\sin(\theta)$$, then $$dx = 5\cos(\theta) d\theta$$

When $$x=0$$, $$5\sin(\theta) = 0$$ so $$\theta = 0$$

When $$x=4$$, $$5\sin(\theta) = 4$$ so $$\theta = \arcsin(\frac{4}{5})$$

Therefore,

$$I_1 = \int_{0}^{\arcsin(\frac{4}{5})} \sqrt{25-25\sin^2(\theta)} (5\cos(\theta)) d\theta$$ $$I_1 = \int_{0}^{\arcsin(\frac{4}{5})} 5\sqrt{1-\sin^2(\theta)} (5\cos(\theta)) d\theta$$ $$I_1 = 25 \int_{0}^{\arcsin(\frac{4}{5})} \cos^2(\theta) d\theta$$

Using the identity $$\cos^2(\theta) = \frac{1+\cos(2\theta)}{2}$$:

$$I_1 = 25 \int_{0}^{\arcsin(\frac{4}{5})} \frac{1+\cos(2\theta)}{2} d\theta$$ $$I_1 = \frac{25}{2} \int_{0}^{\arcsin(\frac{4}{5})} (1+\cos(2\theta)) d\theta$$ $$I_1 = \frac{25}{2} \left[ \theta + \frac{1}{2}\sin(2\theta) \right]_{0}^{\arcsin(\frac{4}{5})}$$

Using the double angle identity, $$\sin(2\theta) = 2\sin(\theta)\cos(\theta)$$:

$$I_1 = \frac{25}{2} \left[ \theta + \sin(\theta)\cos(\theta) \right]_{0}^{\arcsin(\frac{4}{5})}$$

Since $$x = 5\sin(\theta)$$, then $$\sin(\theta) = \frac{x}{5}$$. We also know that $$\theta = \arcsin(\frac{4}{5})$$. Thus, $$\sin(\theta) = \frac{4}{5}$$ and $$\cos(\theta) = \sqrt{1 - \sin^2(\theta)} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$$

$$I_1 = \frac{25}{2} \left[ \arcsin(\frac{4}{5}) + \frac{4}{5} \cdot \frac{3}{5} \right] - 0$$ $$I_1 = \frac{25}{2} \arcsin(\frac{4}{5}) + \frac{25}{2} \cdot \frac{12}{25} = \frac{25}{2} \arcsin(\frac{4}{5}) + 6$$

Evaluating $$I_2$$

$$I_2 = \int_{0}^{4} \frac{25}{2\sqrt{25-x^2}} dx$$ $$I_2 = \frac{25}{2} \int_{0}^{4} \frac{1}{\sqrt{25-x^2}} dx$$ $$I_2 = \frac{25}{2} \left[ \arcsin(\frac{x}{5}) \right]_{0}^{4}$$ $$I_2 = \frac{25}{2} \arcsin(\frac{4}{5}) - \frac{25}{2} \arcsin(0)$$ $$I_2 = \frac{25}{2} \arcsin(\frac{4}{5})$$

Final Calculation

$$I_1 - I_2 = \frac{25}{2} \arcsin(\frac{4}{5}) + 6 - \frac{25}{2} \arcsin(\frac{4}{5})$$ $$I_1 - I_2 = 6$$

Therefore, the value of the definite integral is:

6