Science:Math Exam Resources/Courses/MATH105/April 2011/Question 09 (b)

MATH105 April 2011

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[hide] Question 09 (b)
Each of the short-answer questions below is worth 5 points. Put your answer in the box provided and show your work. No credit will be given for the answer without the correct accompanying work. What is the slope of the curve $y=\int _{0}^{2\sin x}e^{-t^{2}}\,dt$ at the point $x=0?$

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If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

[show] Hint 1
Apply the fundamental theorem of calculus to the integral. i.e: $\int _{a}^{b}f(x)\,dx=F(b)-F(a)$ where F'(x) = f(x).

[show] Hint 2
Consider the first hint. Can you take the derivative of $\int _{a}^{g(x)}f(t)\,dt$ with respect to $x$ where $g$ is a function of $x$ ?

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[show] Solution
To evaluate the slope of the curve at $x=0$ , we let $\displaystyle f(x)=\int _{0}^{2\sin(x)}e^{-t^{2}}\,dt$ and evaluate $f'(0)$ . By the fundamental theorem of calculus, we know that $\displaystyle f(x)=\int _{0}^{2\sin(x)}e^{-t^{2}}\,dt=F(2\sin(x))-F(0)$ where $F$ is the antiderivative of the integrand (i.e: $F'(t)=e^{-t^{2}}$ ). Taking the derivative with respect to $x$ , we find ${\begin{aligned}f'(x)&={\frac {d}{dx}}\left[F(2\sin(x))-F(0)\right]\\&=F'(2\sin(x))\cdot {\frac {d}{dx}}(2\sin(x))\\&=e^{-4\sin ^{2}(x)}\cdot 2\cos(x)\\f'(0)&=2\end{aligned}}$ So the slope of the curve $y=f(x)$ at $x=0$ is 2.