The correct option is D π/4 ∫_0^1√(1 x^2)· #10;dx #10; ∫√(1 x^2)dx=x2√(a^2 x^2)+a^22 #10;sin^ 1 ( xa )+ c #10; ∫_0^1√(1 x^2)· dx= ...

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Integration as Antiderivative

∫01√1-x2dx=

Question

$\int_{0}^{1} \sqrt{1 - x^{2}} d x_{=}$

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 D $\frac{π}{4}$
$\int_{0}^{1} \sqrt{1 - x^{2}} \cdot d x$
$\int \sqrt{1 - x^{2}} d x = \frac{x}{2} \sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2} s i n^{- 1} (\frac{x}{a}) + c$
$\int_{0}^{1} \sqrt{1 - x^{2}} \cdot d x = (\frac{1}{2} \sqrt{1^{2} - x^{2}} + \frac{a^{2}}{2} s i n^{- 1} (\frac{x}{a}) + c) \int_{0}^{1}$
$= (\frac{1}{2} (0) + \frac{1}{2} s i n^{- 1} (1) + c) - (0 + 0 + c)$
$= \frac{1}{2} \frac{π}{2}$
$= \frac{π}{4}$
$\int_{0}^{1} \sqrt{1 - x^{2}} d x = \frac{π}{4}$

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Similar questions

Q. Observe the following Lists

List-I	List-II
A: $\int_{- 2}^{2} \frac{1}{4 + x^{2}} d x$	1) $\frac{π}{3}$
B: $\int_{1}^{2} \frac{1}{x \sqrt{x^{2} - 1}} d x$	2) 0
C: $\int_{0}^{π} cos 3 x . cos 2 x d x$	3) $\frac{π}{4}$
	4) $\frac{π}{2}$

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

Q. If

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

and

y (\frac{π}{4}) = \frac{4}{3}

, then

y (- \frac{π}{4})

equals :

Q. The value of the integral

\frac{1}{\sqrt{2}} \int 0 \frac{{sin}^{- 1} x d x}{(1 - x^{2})^{\frac{3}{2}}}

Solve the equation: $\sqrt{(\frac{1}{16} + c o s^{4} x - \frac{1}{2} c o s^{2} x)} + \sqrt{(\frac{9}{16} + c o s^{4} x - \frac{3}{2} c o s^{2} x)} = \frac{1}{2}$

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∫10√1−x2dx=

The correct option is D π4 ∫10√1−x2⋅dx ∫√1−x2dx=x2√a2−x2+a22sin−1(xa)+c ∫10√1−x2⋅dx=(12√12−x2+a22sin−1(xa)+c)∫10 =(12(0)+12sin−1(1)+c)−(0+0+c) =12π2 =π4 ∫10√1−x2dx=π4

$\int_{0}^{1} \sqrt{1 - x^{2}} d x_{=}$