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JEE Main 2016 · 10 Apr · Shift 1, Mathematics question 75

The value of the integral ∫410[x2] dx[x2−28x+196]+[x2]\displaystyle\int_4^{10}\frac{[x^2]\,dx}{[x^2-28x+196]+[x^2]}, where [x] denotes the greatest integer less than or equal to x, is :

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Answer and worked solution

Official answer: B

x2−28x+196=(14−x)2x^2-28x+196=(14-x)^2, and x→14−xx\to14-x maps [4, 10] onto itself while swapping the two terms. Adding the integral to its transformed form gives 2I=∫410dx=62I=\int_4^{10}dx=6, so I = 3.

Topics:Properties of definite integrals
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