#### Question

If `int_0^h1/(2+8x^2)dx=pi/16 `then find the value of h.

#### Solution

`int_0^h1/(2+8x^2)dx=pi/16`

`therefore 1/8int_0^h1/(x^2+(1/4))dx=pi/16`

`therefore 1/8int_0^h1/(x^2+(1/2)^2)dx=pi/16`

`therefore 1/8 xx 1/(1/2) [tan^-1(x/(1/2))]_0^h=pi/16`

`therefore 1/8 xx 2 [tan^-1(2x)]_0^h=pi/16`

`therefore 1/4tan^-1(2h) - 1/4tan^-1(0)=pi/16`

`therefore 1/4tan^-1(2h)=pi/16`

`therefore tan^-1(2h)=pi/4`

`therefore 2h=tan(pi/4)`

`therefore 2h=1 " ... "[because tan(pi/4)=1]`

`therefore h=1/2`

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#### APPEARS IN

Solution If ∫h0 1/(2+8x2)dx=π/16 then find the value of h. Concept: Fundamental Theorem of Calculus.