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Evaluate the following integral $$\int_{-3}^{3} \sqrt{9-x^{2}} \mathrm{d} x$$
$$\int_{-3}^{3} \sqrt{9-x^{2}} d x=\frac{1}{2} \pi a^{2}$$
$$a^{2}=(3)^{2}=9$$
$$\int_{-3}^{3} \sqrt{9-x^{2}} d x=\frac{1}{2} \pi(9)=\frac{9}{2} \pi$$
Evaluate the following integral $$\int_{-\sqrt{7}}^{0} \sqrt{7-x^{2}} \mathrm{d} x$$
$$\int_{-\sqrt{7}}^{0} \sqrt{7-x^{2}} d x=\left(\frac{1}{2} \pi a^{2}\right) \frac{1}{2}$$
$$=\left(\frac{1}{2} \pi(\sqrt{7})^{2}\right) \frac{1}{2}$$
$$=\frac{1}{4} \pi 7=\frac{7}{4} \pi$$
Use part 1 of the Fundamental Theorem of Calculus to find the derivative of the function $$g(x)=\int_{0}^{x} \sqrt{t+t^{3}} d t$$
$$g(x)=\int_{0}^{x} \sqrt{t+t^{3}} d t=$$
$$g(x)=\sqrt{x+x^{3}}$$
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Evaluate the following integral $$\int_{-3}^{3} \sqrt{9-x^{2}} \mathrm{d} x$$
$$\int_{-3}^{3} \sqrt{9-x^{2}} d x=\frac{1}{2} \pi a^{2}$$
$$a^{2}=(3)^{2}=9$$
$$\int_{-3}^{3} \sqrt{9-x^{2}} d x=\frac{1}{2} \pi(9)=\frac{9}{2} \pi$$
Evaluate the following integral $$\int_{-\sqrt{7}}^{0} \sqrt{7-x^{2}} \mathrm{d} x$$
$$\int_{-\sqrt{7}}^{0} \sqrt{7-x^{2}} d x=\left(\frac{1}{2} \pi a^{2}\right) \frac{1}{2}$$
$$=\left(\frac{1}{2} \pi(\sqrt{7})^{2}\right) \frac{1}{2}$$
$$=\frac{1}{4} \pi 7=\frac{7}{4} \pi$$
Use part 1 of the Fundamental Theorem of Calculus to find the derivative of the function $$g(x)=\int_{0}^{x} \sqrt{t+t^{3}} d t$$
$$g(x)=\int_{0}^{x} \sqrt{t+t^{3}} d t=$$
$$g(x)=\sqrt{x+x^{3}}$$
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