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JEE Mains · Maths · STD 12 - 7.2 definite integral
\(\lim _{x \rightarrow 0} \frac{\int_{0}^{x^{2}}(\sin \sqrt{t}) dt }{x^{3}}\) is equal to
- A \(\frac{2}{3}\)
- B \(\frac{2}{3}\)
- C \(0\)
- D \(\frac{1}{15}\)
Answer & Solution
Correct Answer
(A) \(\frac{2}{3}\)
Step-by-step Solution
Detailed explanation
\(\lim _{x \rightarrow 0^{+}} \frac{\int_{0}^{x^{2}} \sin \sqrt{t} d t}{x^{3}}=\lim _{x \rightarrow 0^{+}} \frac{(\sin x) 2 x}{3 x^{2}}\) \(=\lim _{x \rightarrow 0^{+}}\left(\frac{\sin x}{x}\right) \times \frac{2}{3}=\frac{2}{3}\)
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