Darryl English

2022-07-25

the solid lies between planes perpendicular to the x-axis at x=-1 and x=1. the cross-sections perpendicular to the x-axisare

a. circles whose diameters stretch from the curve $y=-\frac{1}{\sqrt{1+{x}^{2}}}$ to the curve $y=\frac{1}{\sqrt{1+{x}^{2}}}$

b. vertical squares whose base edges run along the same ends

a. circles whose diameters stretch from the curve $y=-\frac{1}{\sqrt{1+{x}^{2}}}$ to the curve $y=\frac{1}{\sqrt{1+{x}^{2}}}$

b. vertical squares whose base edges run along the same ends

suponeriq

Beginner2022-07-26Added 10 answers

Have you had the derivatives of inverse trigonometric functions yet?

$\frac{d}{dx}[{\mathrm{tan}}^{-1}(x)]=\frac{1}{1+{x}^{2}}$

If so you can use the definition of the anti-derivative to deduce

$\frac{d}{dx}[{\mathrm{tan}}^{-1}(x)]=\frac{1}{1+{x}^{2}}\Rightarrow \int \frac{1}{1+{x}^{2}}dx={\mathrm{tan}}^{-1}(x)$

$\frac{d}{dx}[{\mathrm{tan}}^{-1}(x)]=\frac{1}{1+{x}^{2}}$

If so you can use the definition of the anti-derivative to deduce

$\frac{d}{dx}[{\mathrm{tan}}^{-1}(x)]=\frac{1}{1+{x}^{2}}\Rightarrow \int \frac{1}{1+{x}^{2}}dx={\mathrm{tan}}^{-1}(x)$

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