Area bounded by the curves \( y=x^3,y=x+6 \) and \( x=0 \) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 10 Sq. units
The curves meet where
\( \displaystyle x^3=x+6 \)
\( \displaystyle x^3-x-6=0 \)
\( \displaystyle (x-2)(x^2+2x+3)=0 \)
\( x^2+2x+3=(x+1)^2+2 \) is never zero, so the only point of intersection is x = 2, y = 8.
Between x = 0 and x = 2 the line lies above the curve (at x = 0 the line is at 6 and the curve at 0), so
\( \displaystyle \text{Area}=\int_{0}^{2}\left(x+6-x^3\right)dx \)
\( \displaystyle =\left[\frac{x^2}{2}+6x-\frac{x^4}{4}\right]_{0}^{2} \)
\( \displaystyle =2+12-4=10 \)
The area is 10 sq. units.
NCERT Exemplar:
Application of Integrals
