Science:Math Exam Resources/Courses/MATH105/April 2017/Question 06 (b)

MATH105 April 2017

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Question 06 (b)
Evaluate $\lim _{x\to 0}{\frac {1}{x}}\int _{0}^{2017x}{\sqrt {2017+t^{2017}}}dt.$.

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
You do not need to compute the antiderivative of ${\sqrt {2017+t^{2017}}}$ here. Try to rewrite the limit in a way that will allow us to apply the fundamental theorem of calculus.

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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. In order to solve this problem, we introduce the auxiliary function $f(y)=\int _{0}^{y}{\sqrt {2017+t^{2017}}}\,dt$. Then we need to compute $\lim _{x\to 0}{\frac {f(2017x)}{x}}$. Notice that f(0) is zero, since the integral from a to a of any function is zero. So the above limit is equal to $\lim _{x\to 0}{\frac {f(2017x)-f(0)}{x}}.$ By the definition of the derivative, this is $g'(0)$, where $g(x)=f(2017x).$ But we also know from the chain rule that $g'(x)=2017f'(2017x)$ so $g'(0)=2017f'(0)$ It remains to compute $f'(0)$. This can be computed using the fundamental theorem of calculus: we have $f'(0)={\sqrt {2017+0^{2017}}}={\sqrt {2017}}$ so we get that $g'(0)=2017{\sqrt {2017}}=2017^{3/2}$. Answer: $\color {blue}2017^{3/2}$