91Ó°ÊÓ

In this case, we don't need to perform any substitution as the integral has a clear antiderivative.

In order to evaluate the definite integral, we first need to find the antiderivative of the function $$\sqrt{100-x^2}$$. Notice that the function looks like a semicircle with radius 10 and centered at the origin. So we can rewrite the function as: $$\int_{5}^{10} \sqrt{100-x^{2}} d x = \int_{5}^{10} 10\sin^2(t) dt$$, where $$x = 10\cos(t)$$. According to the substitution method, we need to find the differential, dx. Since $$x = 10\cos(t)$$, the differential can be found using differentiation: $$\frac{dx}{dt} = -10\sin(t)$$, which means $$dt=-\frac{1}{x}dx = -\frac{dx}{10\cos(t)}$$. Now, we can substitute the function and the differential into the integral: $$\int_{5}^{10} 10\sin^2(t) dt = \int_{5}^{10} 100\sin^2(t) \frac{dx}{dt} dx = \int_{5}^{10} 100 \sin^2(t) \left(-\frac{dx}{10\cos(t)}\right)= \int_{5}^{10} -10 \sin(t) dx$$

Now we have the new integral as $$\int_{5}^{10} -10\sin(t) dx$$. We can evaluate its antiderivative: $$\int -10\sin(t) dt = -10\int\sin(t) dt = 10\cos(t) + C$$ The antiderivative of the original function is found to be $$10\cos(t) + C$$.

Now we need to apply the limits of integration x = 5 and x = 10 to evaluate the definite integral. Remember that we made a substitution, so we need to find the corresponding limits for t. For $$x=5$$, we have $$t = \cos^{-1}\left(\frac{5}{10}\right)=\cos^{-1}\left(0.5\right)= \frac{\pi}{3}$$. For $$x=10$$, we have $$t = \cos^{-1}\left(\frac{10}{10}\right)=\cos^{-1}(1)= 0$$. Now we can evaluate the definite integral: $$\int_{5}^{10} \sqrt{100-x^{2}} d x = \int_{\frac{\pi}{3}}^{0} -10\sin(t) dt = 10\cos(t) \Big|_{\frac{\pi}{3}}^{0} = 10\cos(0)-10\cos\left(\frac{\pi}{3}\right) = 10\times (1-0.5) = 10\times 0.5$$

Finally, we can calculate the value of the definite integral: $$\int_{5}^{10} \sqrt{100-x^{2}} d x = 10\times 0.5 = 5$$ So, the value of the definite integral is 5.