Question

The area bounded by the curve $y=x^4-2x^3+x^2+3$ with $x-$ axis and ordinates corresponding to the minima of y is:

• A. 1 sq unit
• B. $\frac{91}{30}squnit$
• C. $\frac{30}{9}squnit$
• D. $4squnit$
• E. $\frac{30}{91}squnit$

Answer 1

Answer: (B)

The equation of given curve is $y=x^4-2x^3+x^2+3$ On differentiating w. r. t. $x,$ we get $\frac{dy}{dx}=4x^3-6x^2+2x$ Again differentiating, we get $\frac{d^{2}y}{d{x}^2}=12x^2-12x+2$ Put $\frac{dy}{dx}=0$ for maxima or minima $\Rightarrow$ $2x(2x^2-3x+1)=0$ $\Rightarrow$ $x=0,1,\frac{1}{2}$ $\therefore$ ${\left(\frac{d^{2}y}{d{x}^2}\right)}_{x=0}=2$ $\therefore$ Function is minimum at $x=0$ and ${\left(\frac{d^{2}y}{d{x}^2}\right)}_{x=1}=12-12+2=2$ $\therefore$ Function is minimum at $x=\frac{1}{2}$ . also ${\left(\frac{d^{2}y}{d{x}^2}\right)}_{x=\frac{1}{2}}=-1<0$ $\therefore$ Function is maximum at $x=\frac{1}{2}$ . $\therefore$ Required area $={\int }_0^1(x^4-2x^3+x^2+3)dx$ $={\left[\frac{x^5}{5}-\frac{2x^4}{4}+\frac{x^3}{3}+3x\right]}_0^1$ $=\frac{1}{5}-\frac{1}{2}+\frac{1}{3}+3=\frac{91}{30}squnit$