Definite Integration Question 14
Question 14 - 30 January - Shift 1
$\lim _{x \to 0} \frac{48}{x^{4}} \int_0^{x} \frac{t^{3}}{t^{6}+1} d t$ is equal to
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Answer: 12
Solution:
Formula: L’ Hospital Rule, Limit of a function, Leibnitz Theorem
$ L= 48 \lim _{x \to 0} \frac{\int_0^{x} \frac{t^{3}}{t^{6}+1} d t}{x^{4}}(\frac{0}{0}) $
Applying L’Hospitals Rule and using Leibnitz theorem
$L=48 \lim _{x \to 0} \frac{x^{3}}{x^{6}+1} \times \frac{1}{4 x^{3}}$
$L=12$
