the integral in this exercise converges. evaluate the integral without using a table\n int_{0}^{12} \frac{d…

the integral in this exercise converges. evaluate the integral without using a table\n int_{0}^{12} \frac{d x}{sqrt{144-x^{2}}} \n int_{0}^{12} \frac{d x}{sqrt{144-x^{2}}}=

the integral in this exercise converges. evaluate the integral without using a table\n int_{0}^{12} \frac{d x}{sqrt{144-x^{2}}} \n int_{0}^{12} \frac{d x}{sqrt{144-x^{2}}}=

Answer

Explanation:

Step1: Recall the integral formula

Recall the formula (\int\frac{dx}{\sqrt{a^{2}-x^{2}}}=\sin^{- 1}(\frac{x}{a})+C) ((a>0)). Here (a = 12) since (a^{2}=144).

Step2: Apply the fundamental theorem of calculus

By the fundamental theorem of calculus (\int_{0}^{12}\frac{dx}{\sqrt{144 - x^{2}}}=\left[\sin^{-1}(\frac{x}{12})\right]{0}^{12}). Substitute the upper - limit (x = 12) and lower - limit (x = 0) into (\sin^{-1}(\frac{x}{12})): When (x = 12), (\sin^{-1}(\frac{12}{12})=\sin^{-1}(1)). When (x = 0), (\sin^{-1}(\frac{0}{12})=\sin^{-1}(0)). We know that (\sin^{-1}(1)=\frac{\pi}{2}) and (\sin^{-1}(0)=0). So (\left[\sin^{-1}(\frac{x}{12})\right]{0}^{12}=\sin^{-1}(1)-\sin^{-1}(0)).

Answer:

(\frac{\pi}{2})