Calculus

Application of Integrals

Medium

common

Area bounded by the curves $y=x^3$ , $y=x+6$ and $x=0$ is

20 Sq. units

16 Sq. units

10 Sq. units

8 Sq. units

✅ Correct Option: 3

The curves meet where $x^3=x+6$ , i.e. $x=2$ . Between $x=0$ and $x=2$ the line lies above the cubic, so area $=\int_0^2\left(x+6-x^3\right)dx=\left[\frac{x^2}{2}+6x-\frac{x^4}{4}\right]_0^2=2+12-4=10$ .

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Area bounded by the curves y=x3y=x^3y=x3, y=x+6y=x+6y=x+6 and x=0x=0x=0 is

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CUET Mathematics Past Year Questions (Topic-Wise)

Area bounded by the curves $y=x^3$ , $y=x+6$ and $x=0$ is