Question

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`

Answer 1

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`