Area of the region bounded by the curve y= x, y= x and x-axis, (0≤ x≤()/2) is Option 1: (2+2) Square units Option 2: (2-1) Square units Option 3: (2-2) Square units Option 4: (2+1) Square units

Answer: (2-2) Square units. Solution: The curves meet at x=()/4, and the region lies under x on [0,()/4] and under x on [()/4,()/2]. Area =_0^(/4) x dx+_(/4)^(/2) x dx=2(1-1/(2))=2-2.

Calculus

Application of Integrals

Medium

core

Area of the region bounded by the curve $y=\sin x, y=\cos x$ and $x-$ axis, $\left(0\le x\le\frac{\pi}{2}\right)$ is;

$(2+\sqrt2)$ Square units

$(\sqrt2-1)$ Square units

$(2-\sqrt2)$ Square units

$(\sqrt2+1)$ Square units

✅ Correct Option: 3

The curves meet at $x=\frac{\pi}{4}$ , and the region lies under $\sin x$ on $\left[0,\frac{\pi}{4}\right]$ and under $\cos x$ on $\left[\frac{\pi}{4},\frac{\pi}{2}\right]$ .

Area $=\int_0^{\pi/4}\sin x\,dx+\int_{\pi/4}^{\pi/2}\cos x\,dx=2\left(1-\frac{1}{\sqrt2}\right)=2-\sqrt2$ .

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The area bounded by $y = 3x + 1$ , $x = 0$ , $y = 0$ and $x = a$ is 8 Sq.units. Then value of $a$ (where $a > 0$ ) is

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The area bounded by the curve $y = 4 + 3x - x^2$ and $x$ -axis is equal to

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The area of the region bounded by parabola $x^2 = 4y$ , straight line $x = 2$ and $x$ -axis, is

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The area of the region bounded by the curve $x^2=4y$ , the line $x=4y-2$ and x-axis is

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Area of the region bounded by the curve $y = \sin x$ and x-axis between $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$ is

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The area bounded by the x-axis and the parabola $y = 3x-x^2$ is:

Area of the region bounded by the curve y=sin⁡x,y=cos⁡xy=\sin x, y=\cos xy=sinx,y=cosx and x−x-x−axis, (0≤x≤π2)\left(0\le x\le\frac{\pi}{2}\right)(0≤x≤2π​) is;

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CUET Mathematics Past Year Questions (Topic-Wise)

Area of the region bounded by the curve $y=\sin x, y=\cos x$ and $x-$ axis, $\left(0\le x\le\frac{\pi}{2}\right)$ is;