Science:Math Exam Resources/Courses/MATH103/April 2016/Question 01 (d)

MATH103 April 2016

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Question 01 (d)
Compute $\displaystyle {\frac {d}{dx}}\left(\int _{0}^{3x}\ e^{\cos t}dt\right)$ at $x=0$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Write the given integral as a composition of two functions. Then, using the chain rule and the Fundamental theorem of Calculus, find the derivative at 0.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Recall that part 1 of the Fundamental Theorem of Calculus is ${\frac {d}{dy}}\int _{a}^{y}f(t)dt=f(y)$. First, we can write the given integral as a composite of two functions: $\int _{0}^{3x}e^{\cos(t)}dt=F(u(x))$ where $F(u)=\int _{0}^{u}e^{\cos(t)}dt$ and $u(x)=3x$.. So, by the chain rule, we have ${\frac {d}{dx}}\int _{0}^{3x}e^{\cos(t)}dt={\frac {d}{dx}}F(u(x))=F'(u(x))u'(x).$ The Fundamental Theorem of Calculus implies that $F'(u(x))={\frac {d}{du}}\int _{0}^{u}e^{\cos(t)}dt{\bigg |}_{u=u(x)}=e^{\cos(u)}{\bigg |}_{u=u(x)}=e^{\cos(3x)};$ and $u'(x)=3.$ We finally obtain $\left.{\frac {d}{dx}}\int _{0}^{3x}e^{\cos(t)}dt\right|_{x=0}=F'(u(0))u'(0)=3e^{\cos(0)}=\color {blue}3e.$