The area enclosed by the graph of the functio | vedclass

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

Question

(B) The function is $y = \ln^2 x - 1$. The $4^{th}$ quadrant corresponds to the region where $y  0$.Setting $y = 0$,we get $\ln^2 x = 1$,

Vedclass · Accepted Answer

$\frac{4}{e}$

Vedclass · Answer

(B) The function is $y = \ln^2 x - 1$. The $4^{th}$ quadrant corresponds to the region where $y  0$.Setting $y = 0$,we get $\ln^2 x = 1$,so $\ln x = \pm 1$,which gives $x = e$ or $x = 1/e$.For $x \in (1/e, e)$,$\ln x$ lies between $-1$ and $1$,so $\ln^2 x The area $A$ is given by $\int_{1/e}^{e} |\ln^2 x - 1| dx = \int_{1/e}^{e} (1 - \ln^2 x) dx$.Using integration by parts for $\int \ln^2 x dx = x \ln^2 x - 2x \ln x + 2x$,we have:$\int (1 - \ln^2 x) dx = x - (x \ln^2 x - 2x \ln x + 2x) = -x \ln^2 x + 2x \ln x - x$.Evaluating from $1/e$ to $e$:At $x = e$: $-e(1)^2 + 2e(1) - e = -e + 2e - e = 0$.At $x = 1/e$: $-(1/e)(-1)^2 + 2(1/e)(-1) - (1/e) = -1/e - 2/e - 1/e = -4/e$.Thus,$A = 0 - (-4/e) = 4/e$.

The area enclosed by the graph of the function $y = \ln^2 x - 1$ lying in the $4^{th}$ quadrant is:

Similar Questions

The area inside the parabola $y^2 = 4ax$,between the lines $x = a$ and $x = 4a$ is equal to

The area (in square units) of the region bounded by $x^2=8y$, $x=4$ and the $X$-axis is:

The area bounded by the curve $y^2 = x$,the line $y = 4$,and the $y$-axis is

If the line $y=mx$ bisects the area enclosed by the lines $x=0, y=0, x=\frac{3}{2}$ and the curve $y=1+4x-x^2$,then $12m$ is equal to ..... .

The area of the region bounded by the parabola $y=x^2$ and the line $y=4$ is . . . . . . sq. units.

Vedclass Test Series

Exam Paper Generator

Online Exam Module