The area bounded by a curve,the x-axis,and th | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let the curve be $y = f(x)$. The area $A$ bounded by the curve,the $x$-axis,and the ordinate at $x$ is given by $A = \int_0^x y \, dx$. The arc le

Vedclass · Accepted Answer

$(B)$ and $(C)$ both

Vedclass · Answer

(D) Let the curve be $y = f(x)$. The area $A$ bounded by the curve,the $x$-axis,and the ordinate at $x$ is given by $A = \int_0^x y \, dx$. The arc length $s$ of the curve from $x=0$ to $x$ is given by $s = \int_0^x \sqrt{1 + (y')^2} \, dx$. According to the problem,$A = s$,so $\int_0^x y \, dx = \int_0^x \sqrt{1 + (y')^2} \, dx$. Differentiating both sides with respect to $x$ using the Fundamental Theorem of Calculus,we get $y = \sqrt{1 + (y')^2}$. Squaring both sides,$y^2 = 1 + (y')^2$,which implies $(y')^2 = y^2 - 1$. Thus,$y' = \pm \sqrt{y^2 - 1}$. Separating variables,$\frac{dy}{\sqrt{y^2 - 1}} = \pm dx$. Integrating both sides,$\ln|y + \sqrt{y^2 - 1}| = \pm x + C$. Given the curve passes through $(0, 1)$,at $x=0$,$y=1$. Substituting these values: $\ln|1 + \sqrt{1-1}| = \pm 0 + C \implies \ln(1) = C \implies C = 0$. So,$\ln|y + \sqrt{y^2 - 1}| = x$ or $-x$. This simplifies to $y + \sqrt{y^2 - 1} = e^x$ or $y + \sqrt{y^2 - 1} = e^{-x}$. Solving for $y$,we get $y = \frac{e^x + e^{-x}}{2} = \cosh(x)$. Also,if $y=1$,then $y'=0$,which satisfies $y = \sqrt{1+(y')^2} \implies 1 = \sqrt{1+0} = 1$. Thus,both $y = \cosh(x)$ and $y=1$ are solutions. Therefore,the correct option is $(D)$.

The area bounded by a curve,the $x$-axis,and the ordinate of some point $(x, y)$ of the curve is equal to the length of the corresponding arc of the curve from a fixed point. If the curve passes through the point $P(0, 1)$,then the equation of this curve is:

Similar Questions

Let $f:[0,2] \rightarrow R$ be a function which is continuous on $[0,2]$ and is differentiable on $(0,2)$ with $f(0)=1$. Let $F(x)=\int_0^{x^2} f(\sqrt{t}) dt$ for $x \in [0,2]$. If $F'(x)=f'(x)$ for all $x \in (0,2)$,then $F(2)$ equals

If $y = At^2 + \frac{B}{t}$ ($A, B$ are parameters) is the general solution of the differential equation $f(t) y''(t) + g(t) y'(t) + h(t) y = 0$,then $2 f(t) + t^2 h(t) =$

The number of solutions of $\frac{dy}{dx} = \frac{y+1}{x-1}$,when $y(1) = 2$ is

An ice ball melts at a rate proportional to the amount of ice present at that instant. Half of the initial quantity of ice melts in $15 \text{ minutes}$. Let $x_0$ be the initial quantity of ice. If after $30 \text{ minutes}$ the amount of ice left is $k x_0$,then the value of $k$ is:

$A$ curve passes through the point $(3,2)$ for which the segment of the tangent line contained between the coordinate axes is bisected at the point of contact. The equation of the curve is

Vedclass Test Series

Exam Paper Generator

Online Exam Module