The area (in sq. units) of the region bounded | vedclass

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

Question

(D) To find the area bounded by the curves $y = x \log x$ and $y = 2x - 2x^2$,we first find their points of intersection by setting $x \log x = 2x - 2

Vedclass · Accepted Answer

$\frac{7}{12}$

Vedclass · Answer

(D) To find the area bounded by the curves $y = x \log x$ and $y = 2x - 2x^2$,we first find their points of intersection by setting $x \log x = 2x - 2x^2$. For $x > 0$,we divide by $x$: $\log x = 2 - 2x$,which implies $\log x + 2x - 2 = 0$. By inspection,$x = 1$ is a solution since $\log(1) + 2(1) - 2 = 0 + 2 - 2 = 0$. For $x \in (0, 1]$,the curve $y = 2x - 2x^2$ lies above $y = x \log x$. The area $A$ is given by the integral $\int_{0}^{1} (2x - 2x^2 - x \log x) dx$. Evaluating the integral: $\int_{0}^{1} 2x dx = [x^2]_{0}^{1} = 1$. $\int_{0}^{1} 2x^2 dx = [\frac{2}{3}x^3]_{0}^{1} = \frac{2}{3}$. Using integration by parts for $\int x \log x dx$: $\int x \log x dx = \frac{x^2}{2} \log x - \int \frac{x^2}{2} \cdot \frac{1}{x} dx = \frac{x^2}{2} \log x - \frac{x^2}{4}$. Evaluating from $0$ to $1$: $[\frac{x^2}{2} \log x - \frac{x^2}{4}]_{0}^{1} = (0 - \frac{1}{4}) - (0) = -\frac{1}{4}$. Thus,$A = 1 - \frac{2}{3} - (-\frac{1}{4}) = 1 - \frac{2}{3} + \frac{1}{4} = \frac{12 - 8 + 3}{12} = \frac{7}{12}$.

The area (in sq. units) of the region bounded by the curves $y = x \log x$ and $y = 2x - 2x^2$ is

Similar Questions

If $A$ is the area in the first quadrant enclosed by the curve $C: 2x^2 - y + 1 = 0$,the tangent to $C$ at the point $(1, 3)$ and the line $x + y = 1$,then the value of $60A$ is

The area of the plane region bounded by the curves $x + 2y^2 = 0$ and $x + 3y^2 = 1$ is equal to

The parabola $y^2=4x+1$ divides the disc $x^2+y^2 \leq 1$ into two regions with areas $A_1$ and $A_2$. Then,$|A_1-A_2|$ equals

Let the area enclosed by the lines $x + y = 2, y = 0, x = 0$ and the curve $f(x) = \min \left\{x^2 + \frac{3}{4}, 1 + [x]\right\}$,where $[x]$ denotes the greatest integer $\leq x$,be $A$. Then the value of $12A$ is $............$.

Let $R$ denote the set of all real numbers. Then the area of the region $\{(x, y) \in R \times R : x > 0, y > \frac{1}{x}, 5x - 4y - 1 > 0, 4x + 4y - 17 < 0\}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module