Area bounded by the curves y = ^2 x, x = 0, y = 0 and x =, is: Option 1: Square units Option 2: ()/2 Square units Option 3: ()/4 Square units Option 4: (()/2 - 1/2) Square units

Answer: ()/2 Square units. Solution: Since ^2 x≥ 0 throughout, the area is _0^ ^2x dx=_0^(1+ 2x)/2 dx. The 2x part integrates to zero over [0,], leaving ()/2 square units.

Calculus

Application of Integrals

Easy

core

Area bounded by the curves $y = \cos^2 x, x = 0, y = 0$ and $x = \pi$ , is:

$\pi$ Square units

$\frac{\pi}{2}$ Square units

$\frac{\pi}{4}$ Square units

$\left(\frac{\pi}{2} - \frac{1}{2}\right)$ Square units

✅ Correct Option: 2

Since $\cos^2 x\ge 0$ throughout, the area is $\int_0^\pi \cos^2x\,dx=\int_0^\pi\frac{1+\cos 2x}{2}\,dx$ . The $\cos 2x$ part integrates to zero over $[0,\pi]$ , leaving $\frac{\pi}{2}$ square units.

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Area bounded by the curves y=cos⁡2x,x=0,y=0y = \cos^2 x, x = 0, y = 0y=cos2x,x=0,y=0 and x=πx = \pix=π, is:

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CUET Mathematics Past Year Questions (Topic-Wise)

Area bounded by the curves $y = \cos^2 x, x = 0, y = 0$ and $x = \pi$ , is: