Question
If $ \displaystyle f(x) = \int^{\sin x}_0 \sqrt{1 + t^2} \, dt $ and $ \displaystyle g(y) = \int^y_3 f(x) \, dx $, find $ g''(\pi/6) $.
Step 1
Step 1: f(x)=∫_0^{sin x} √(1+t^2) dt ⇒ f'(x)=√(1+sin^2 x)·cos x by FTC and chain rule. Show more…
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$$ \begin{array}{l}{\text { If } f(x)=\int_{0}^{\sin x} \sqrt{1+t^{2}} d t \operatorname{and} g(y)=\int_{3}^{y} f(x) d x} \\ {\text { find } g^{\prime \prime}(\pi / 6) .}\end{array} $$
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If $f(x)=\int_{0}^{\sin x} \sqrt{1+t^{2}} d t$ and $g(y)=\int_{3}^{y} f(x) d x$, find $g^{\prime \prime}(\pi / 6) .$
If $f(x)=\int_{0}^{\sin x} \sqrt{1+t^{2}} d t$ and $g(y)=\int_{3}^{y} f(x) d x,$ find $g^{-2}(\pi / 6)$
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