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$\int_{0}^{10} \frac{x^{10}}{(10-x)^{10}+x^{10}} d x$ is equal to
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2717 Upvotes
Verified Answer
The correct answer is:
5
Let $I=\int_{0}^{10} \frac{x^{10}}{(10-x)^{10}+x^{10}} d x$...(1)
$I=\int_{0}^{10} \frac{(10-x)^{10}}{(10-x)^{10}+x^{10}} d x$...(2)
Adding
(1) and (2), we get
$2 I=\int_{0}^{10} d x \Rightarrow 2 I=10 \Rightarrow I=5$
$I=\int_{0}^{10} \frac{(10-x)^{10}}{(10-x)^{10}+x^{10}} d x$...(2)
Adding
(1) and (2), we get
$2 I=\int_{0}^{10} d x \Rightarrow 2 I=10 \Rightarrow I=5$
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