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प्रश्न
Evaluate the following:
`int_0^1 (x"d"x)/sqrt(1 + x^2)`
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उत्तर
Let I = `int_0^1 (x"d"x)/sqrt(1 + x^2)`
Put 1 + x2 = t
⇒ 2x dx = dt
⇒ x dx = `"dt"/2`
Changing the limits, we have
When x = 0
∴ t = 1
When x = 1
∴ t = 2
∴ I = `1/2 int_1^2 "dt"/sqrt("t")`
= `1/2 * 2["t"^(1/2)]_1^2`
= `sqrt(2) - 1`
Hence, I = `sqrt(2) - 1`.
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