JEE maths#23

Calculus Level 2

If ∫ 0 10 f ( x ) d x = 5 \displaystyle\int_{0}^{10} f(x) dx = 5 and ∑ k = 1 10 ∫ 0 1 f ( k − 1 + x ) d x = R . \displaystyle\sum_{k=1}^{10} \int_{0}^{1} f(k-1+x) dx = R.

Then find the value of R R .

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The answer is 5.

Rahil Sehgal by Rahil Sehgal , 20, India

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2 solutions

Relevant wiki: Basic Properties of Integrals

With a suitable change of variables y = x + k − 1 y=x+k-1 ,

∑ k = 0 9 ∫ 0 1 f ( x + k ) d x = ∑ k = 0 9 ∫ k k + 1 f ( x ) d x \displaystyle \sum\limits_{k=0}^9 \int\limits_0^1 \; f(x+k)\; dx=\sum\limits_{k=0}^9 \int\limits_k^{k+1} \; f(x)\; dx

R = ∑ k = 0 9 ∫ k k + 1 f ( x ) d x = ( ∫ 0 1 + ∫ 1 2 + ∫ 2 3 + ⋯ + ∫ 9 10 ) f ( x ) d x = ∫ 0 10 f ( x ) d x = 5 \displaystyle \begin{aligned} R&=\sum\limits_{k=0}^9 \int\limits_k^{k+1} \; f(x)\; dx \\ &=\left(\int_0^1+\int_1^2+\int_2^3+\cdots+\int_9^{10} \right)f(x)\; dx \\ &=\int_0^{10} \; f(x)\;dx \\&=\boxed{5}\end{aligned}

7 Helpful 0 Interesting 0 Brilliant 0 Confused
Aakash Khandelwal
Apr 26, 2017

100% got it right..... amazing

0 Helpful 0 Interesting 0 Brilliant 0 Confused

now it is 98% still amazing..

Rahil Sehgal - 4 years, 1 month ago

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