Question

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Answer 1

✪ᴀɴsᴡᴇʀ✪

$\boxed{\green{\int_{2}^{4} \frac{\sqrt{x} } {\sqrt{6-x} +\sqrt{x}}dx = 1} }$

★ᴇxᴘʟᴀɪɴᴀᴛɪᴏɴ★

᯽ғᴏʀᴍᴜʟᴀ᯽

$\int_{a}^{b} f(x)dx = \int_{a}^{b} f((a+b) -x)dx$

☆ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴ☆

$I = \int_{2}^{4} \frac{\sqrt{x} } {\sqrt{6-x} +\sqrt{x}}dx....(1)$

Using the above formula,

$I = \int_{2}^{4} \frac{\sqrt{(2+4)-x} } {\sqrt{6-[(2+4)-x] } +\sqrt{(2+4)-x}}dx$

$I = \int_{2}^{4} \frac{\sqrt{6-x} } {\sqrt{x} +\sqrt{6-x}}dx....(2)$

Adding (1) and (2), we get

$2I = \int_{2}^{4} \frac{\sqrt{x} } {\sqrt{6-x} +\sqrt{x}}dx + \int_{2}^{4} \frac{\sqrt{6-x} } {\sqrt{x} +\sqrt{6-x}}dx$

$2I = \int_{2}^{4} \frac{\sqrt{x} + \sqrt{6-x}} {\sqrt{6-x} +\sqrt{x}}dx$

$2I = \int_{2}^{4} 1dx$

$2I = [x]_{2}^{4}$

$2I = [4-2]$

$2I = 2$

$I = 1$

Answer 2

CRM

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