The area of region bounded by the curves 4y = x^3, y = 0 and x = 4 in the first quadrant is : Option 1: 16 sq units Option 2: 4 sq units Option 3: 32 sq units Option 4: 8 sq units

Answer: 16 sq units. Solution: The curve is y=x^3/4, which lies above the x-axis for x>0. Area =_0^4 x^3/4dx=[x^4/16]_0^4=256/16=16 sq units.

Calculus

Application of Integrals

Easy

common

The area of region bounded by the curves $4y = x^3, y = 0$ and $x = 4$ in the first quadrant is :

16 sq units

4 sq units

32 sq units

8 sq units

✅ Correct Option: 1

The curve is $y=\dfrac{x^3}{4}$ , which lies above the $x$ -axis for $x>0$ . Area $=\displaystyle\int_0^4 \dfrac{x^3}{4}dx=\left[\dfrac{x^4}{16}\right]_0^4=\dfrac{256}{16}=16$ sq units.

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The area of region bounded by the curves 4y=x3,y=04y = x^3, y = 04y=x3,y=0 and x=4x = 4x=4 in the first quadrant is :

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The area of region bounded by the curves $4y = x^3, y = 0$ and $x = 4$ in the first quadrant is :