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Evaluate: xxxdx∫2810-xx+10-xdx - Mathematics

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प्रश्न

Evaluate:

`int_2^8 (sqrt(10 - "x"))/(sqrt"x" + sqrt(10 - "x")) "dx"`

योग

उत्तर

Given, `int_2^8 (sqrt(10 - "x"))/(sqrt"x" + sqrt(10 - "x")) "dx"`

Let I = `int_2^8 (sqrt(10 - "x"))/(sqrt"x" + sqrt(10 - "x")) "dx"` ............(i)

Then using property:

`int_"a"^"b" "f"("x") "dx" = int_"a"^"b" "f"("a" + "b" - "x") "dx"`

I = `int_2^8 (sqrt(10 - (2 + 8 - "x")))/(sqrt(2 + 8 - "x") + sqrt(10 - (2 + 8 - "x"))) "dx"`

= `int_2^8 (sqrt"x")/(sqrt(10 - "x") + sqrt"x") "dx"` ...........(ii)

Adding equation (i) and (ii), we get

2I = `int_2^8 (sqrt(10 - "x") + sqrt"x")/(sqrt"x" + sqrt(10 - "x")) "dx"`

⇒ 2I = `int_2^8 1. "dx"`

⇒ 2I = `["x"]_2^8`

⇒ 2I = 8 − 2

⇒ 2I = 6

∴ I = 3

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2021-2022 (April) Set 1

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