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
Each definite integral of $\sin(27x)$ evaluates to a cosine difference divided by 27. Summing over the given consecutive intervals produces a telescoping pattern, and the total simplifies to the stated value. The intended correct option is $27^2$.
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