The correct option is A 1 π/4 ∫_0^1x^2dx1+x^2 ∫_0^11dx ∫_0^111+x^2dx =1 [ tan^ 1x ] |_0^1 =1 [ tan^ 1 #160; 1 ] =1 π4 Th ...

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First Fundamental Theorem of Calculus

∫01x2/1+x2dx ...

Question

$\int_{0}^{1} \frac{x^{2}}{1 + x^{2}} d x$ equals

1 - \frac{π}{4}

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1 - \frac{π}{3}

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\frac{π}{3}

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\frac{π}{4}

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Solution

The correct option is A $1 - \frac{π}{4}$
$\int_{0}^{1} \frac{x^{2} d x}{1 + x^{2}}$
$\int_{0}^{1} 1 d x - \int_{0}^{1} \frac{1}{1 + x^{2}} d x$
$= 1 - [{tan}^{- 1} x] |_{0}^{1}$
$= 1 - [{tan}^{- 1} 1]$
$= 1 - \frac{π}{4}$
Thus $\int_{0}^{1} \frac{x^{2}}{1 + x^{2}} d x = 1 - \frac{π}{4}$

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Similar questions

\frac{d y}{d x} + \frac{3}{{c o s}^{2} x} = \frac{1}{{c o s}^{2} x}, x ϵ (\frac{- π}{3}, \frac{π}{3}),

y (\frac{π}{4}) = \frac{4}{3},

y (- \frac{π}{4})

\frac{d y}{d x} + \frac{3}{{cos}^{2} x} y = \frac{1}{{cos}^{2} x}, x \in (\frac{- π}{3}, \frac{π}{3})

y (\frac{π}{4}) = \frac{4}{3}

y (- \frac{π}{4})

\frac{d y}{d x} + \frac{3}{{cos}^{2} x} y = \frac{1}{{cos}^{2} x}, x \in (\frac{- π}{3}, \frac{π}{3}),

y (\frac{π}{4}) = \frac{4}{3}

y (- \frac{π}{4})

\int_{0}^{1} s i n^{- 1} (\frac{2 x}{1 + x^{2}}) d x =

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