Science:Math Exam Resources/Courses/MATH101/April 2016/Question 03 (a)

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MATH101 April 2016

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Question 03 (a)
If $F(x)$ is defined by $\displaystyle F(x)=\int _{x^{4}-x^{3}}^{x}e^{\sin t}\,dt$ , find $F'(x)$ .

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
Use the Fundamental Theorem of Calculus: If $G(x)=\int _{a}^{x}g(t)dt,$ then $G'(x)=g(x)$ .

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Solution
First we rewrite the function $F$ as $F(x)=\int _{x^{4}-x^{3}}^{x}e^{\sin t}\,dt=\int _{0}^{x}e^{\sin t}\,dt+\int _{x^{4}-x^{3}}^{0}e^{\sin t}\,dt=\int _{0}^{x}e^{\sin t}\,dt-\int _{0}^{x^{4}-x^{3}}e^{\sin t}\,dt.$ Let $G(x)=\int _{0}^{x}e^{\sin t}\,dt$ . Then, we have $F(x)=G(x)-G(x^{4}-x^{3})$ . and by the Fundamental Theorem of Calculus, $G'(x)={\frac {d}{dx}}\left(\int _{0}^{x}e^{\sin t}\,dt\right)=e^{\sin x},$ Now let's compute the derivative of $F$ : ${\begin{aligned}F'(x)&={\frac {d}{dx}}(G(x)-G(x^{4}-x^{3}))=G'(x)-G'(x^{4}-x^{3}){\frac {d}{dx}}(x^{4}-x^{3})=G'(x)-G'(x^{4}-x^{3})\cdot (4x^{3}-3x^{2})=e^{\sin x}-e^{\sin(x^{4}-x^{3})}\left(4x^{3}-3x^{2}\right).\end{aligned}}$ Here, the second equality follows from the chain rule. Therefore, the answer is $F'(x)=e^{\sin x}-e^{\sin(x^{4}-x^{3})}\left(4x^{3}-3x^{2}\right).$