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The area bounded by the curves {y}={{}^{{-{1}}}{|{{{x}}}|}}{{a

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The area bounded by the curves $y = sin^{- 1} ∣ sin x ∣ and y = (sin^{- 1} ∣ sin x ∣)^{2}, where 0 \leq x \leq 2 π$ , is

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The area bounded by the curves

y = sin^{- 1} ∣ sin x ∣ and y = (sin^{- 1} ∣ sin x ∣)^{2}, where 0 \leq x \leq 2 π

, is

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Practice questions from similar books

Question 1

Find the area bounded by

y = - x^{3} + x^{2} + 16 x and y = 4 x

View solution

Question 2

The area bounded by the curves

y = sin^{- 1} ∣ sin x ∣ and y = (sin^{- 1} ∣ sin x ∣)^{2}, where 0 \leq x \leq 2 π

, is

View solution

Question 3

If y=f(x) is a monotonic function in (a,b), then the area bounded by the ordinates at

x = a, x = b, y = f (x) and y = f (c) (where c \in (a, b)) is minimum when c = \frac{a + b}{2}

Proof : A = a \int c (f (c) - f (x)) d x + c \int b (f (c)) d x

= f (c) (c - a) - a \int c (f (x)) d x + a \int b (f (x)) d x - f (c) (b - c)

\Rightarrow A = [2 c - (a + b)] f (c) + c \int b (f (x)) d x - a \int c (f (x)) d x

Differentiating w.r.t. c, we get

\frac{d A}{d c} = [2 c - (a + b)] f^{'} (c) + 2 f (c) + 0 - f (c) - (f (c) - 0)

For maxima and minima ,

\frac{d A}{d c} = 0

\Rightarrow f^{'} (c) [2 c - (a + b)] = 0 (a s f^{'} (c) \neq = 0)

Hence,

c = \frac{a + b}{2}

Also for c < \frac{a + b}{2}, \frac{d A}{d c} < 0 and for c > \frac{a + b}{2}, \frac{d A}{d c} > 0

Hence, A is minimum when

c = \frac{a + b}{2}

.
If the area bounded by

f (x) = \frac{x ^{3}}{3} - x^{2} + a

and the straight lines x=0, x=2, and the x-axis is minimum, then the value of a is

View solution

Question 4

Computing area with parametrically represented boundaries : If the boundary of a figure is represented by parametric equation, i.e.,

x = x (t), y = (t),

then the area of the figure is evaluated by one of the three formulas :

S = - α \int β y (t) x^{'} (t) d t,

S = α \int β x (t) y^{'} (t) d t,

S = \frac{1}{2} α \int β (x y^{'} - y x^{'}) d t,

Where

α and β

are the values of the parameter t corresponding respectively to the beginning and the end of the traversal of the curve corresponding to increasing t.
The area of the region bounded by an are of the cycloid

x = a (t - sin t), y = a (1 - cos t)

and the x-axis is

View solution

Question Text

The area bounded by the curves

y = sin^{- 1} ∣ sin x ∣ and y = (sin^{- 1} ∣ sin x ∣)^{2}, where 0 \leq x \leq 2 π

, is

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The area bounded by the curves y=sin−1∣sinx∣andy=(sin−1∣sinx∣)2, where 0≤x≤2π, is

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Practice questions from similar books

The area bounded by the curves $y = sin^{- 1} ∣ sin x ∣ and y = (sin^{- 1} ∣ sin x ∣)^{2}, where 0 \leq x \leq 2 π$ , is