Science:Math Exam Resources/Courses/MATH105/April 2013/Question 01 (f)

MATH105 April 2013

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Question 01 (f)
Short-Answer Questions. Put your answer in the box provided but show your work also. Each question is worth 3 marks, but not all questions are of equal difficulty. If $\displaystyle F(x)=\int _{0}^{\cos(x)}{\frac {1}{t^{3}+6}}\,dt$ , find $\displaystyle F'(\pi )$.

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 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.

Hint 1
It is not necessary to calculate $\displaystyle F(x)$. Assume that we had the antiderivative $\displaystyle A(t)$ of $\displaystyle {\frac {1}{t^{3}+6}}$. Then we also know that $\displaystyle A'(t)={\frac {1}{t^{3}+6}}$ Use the function $\displaystyle A(t)$ to express $\displaystyle F(x)$.

Hint 2
Once, you used $\displaystyle A(t)$, we find $\displaystyle F(x)=A(\cos(x))-A(0)$ Do not forget to apply the chain rule by differentiating $\displaystyle F(x)$!

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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. Since we don't want to calculate $\displaystyle F(x)$, we define the antiderivative $\displaystyle A(t)$ of $\displaystyle {\frac {1}{t^{3}+6}}$. This also means that $\displaystyle A'(t)={\frac {1}{t^{3}+6}}$. We use $\displaystyle A(t)$ to express $\displaystyle F(x)$: ${\begin{aligned}F(x)=\int _{0}^{\cos(x)}{\frac {1}{t^{3}+6}}dt=\left[A(t)\right]_{0}^{\cos(x)}=A(\cos(x))-A(0)\end{aligned}}$ Now we can differentiate $\displaystyle F(x)$ by using the chain rule $\displaystyle F'(x)=A'(\cos(x))\cdot (-\sin(x))$ Note that $\displaystyle A(0)$ is a constant for $\displaystyle x$ and cancels by differentiating. We use that we know $\displaystyle A'(t)$, the derivative of $\displaystyle A(t)$, and find $\displaystyle F'(x)={\frac {1}{(\cos x)^{3}+6}}(-\sin(x))=-{\frac {\sin(x)}{(\cos x)^{3}+6}}$ Then, we apply $\displaystyle \pi$, $\displaystyle F'(\pi )=-{\frac {\sin(\pi )}{(\cos \pi )^{3}+6}}=0$

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