Exams › JEE Main › Maths
Evaluate the expression \[\sum_{n=1}^{10} \int_{-2n}^{-n} \sin(27x)\,dx + \sum_{n=1}^{10} \int_{2n-1}^{2n} \sin(27x)\,dx\].
- 27^2
- −54
- 54
- 0
Correct answer: 27^2
Solution
The expression consists of two sums of integrals of the sine function over specific intervals. The first sum evaluates to zero due to the symmetry of the sine function over the intervals, while the second sum results in a positive value, specifically 27^2, due to the periodic nature of the sine function and the properties of definite integrals.
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