$$\mathop {\lim }\limits_{x \to 0} {{\int_0^x {t\sin \left( {10t} \right)dt} } \over x}$$ is equal to
Solution
$$\mathop {\lim }\limits_{x \to 0} {{\int_0^x {t\sin \left( {10t} \right)dt} } \over x}$$
<br><br>This is in ${0 \over 0}$ form.
<br><br>So apply newton leibniz rule
<br><br>$\mathop {\lim }\limits_{x \to 0} {{x.\sin \left( {10x} \right) - 0} \over 1}$ = 0
About this question
Subject: Mathematics · Chapter: Definite Integration · Topic: Properties of Definite Integrals
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