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प्रश्न
By completing the following activity, Evaluate `int_2^5 (sqrt(x))/(sqrt(x) + sqrt(7 - x)) "d"x`.
Solution: Let I = `int_2^5 (sqrt(x))/(sqrt(x) + sqrt(7 - x)) "d"x` ......(i)
Using the property, `int_"a"^"b" "f"(x) "d"x = int_"a"^"b" "f"("a" + "b" - x) "d"x`, we get
I = `int_2^5 ("( )")/(sqrt(7 - x) + "( )") "d"x` ......(ii)
Adding equations (i) and (ii), we get
2I = `int_2^5 (sqrt(x))/(sqrt(x) - sqrt(7 - x)) "d"x + ( ) "d"x`
2I = `int_2^5 (("( )" + "( )")/("( )" + "( )")) "d"x`
2I = `square`
∴ I = `square`
उत्तर
Let I = `int_2^5 (sqrt(x))/(sqrt(x) + sqrt(7 - x)) "d"x` ......(i)
Using the property, `int_"a"^"b" "f"(x) "d"x = int_"a"^"b" "f"("a" + "b" - x) "d"x`, we get
I = `int_2^5 ((sqrt(7 - x)))/(sqrt(7 - x) + (sqrt(x)) "d"x` ......(ii)
Adding equations (i) and (ii), we get
∴ 2I = `int_2^5 (sqrt(x))/(sqrt(x) - sqrt(7 - x)) "d"x + (int_2^5 (sqrt(7 - x))/(sqrt(7 - x) + sqrt(x)) ) "d"x`
∴ 2I = `int_2^5 (sqrt(x) + sqrt(7 - x))/(sqrt(x) + sqrt(7 - x)) "d"x`
∴ 2I = `int_2^5 1*"d"x`
∴ 2I = `[x]_2^5`
∴ 2I = 5 – 2
∴ I = `3/2`
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