Let f(alpha) denote the area of the region in | vedclass

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

Question

(C) For $\alpha = 0$, the function becomes $y = |0 \cdot x - 5| - |1 - 0 \cdot x| + 0 \cdot x^2 = |-5| - |1| + 0 = 5 - 1 = 4$.Thus, $f(0)$ is the area

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(C) For $\alpha = 0$, the function becomes $y = |0 \cdot x - 5| - |1 - 0 \cdot x| + 0 \cdot x^2 = |-5| - |1| + 0 = 5 - 1 = 4$.Thus, $f(0)$ is the area bounded by $x=0, x=1, y^2=x$ and $y=4$. Since $y^2=x$ implies $y=\sqrt{x}$ in the first quadrant, the area is:$f(0) = \int_0^1 (4 - \sqrt{x}) dx = [4x - \frac{2}{3}x^{3/2}]_0^1 = 4 - \frac{2}{3} = \frac{10}{3}$.For $\alpha = 1$, the function becomes $y = |x - 5| - |1 - x| + x^2$. For $x \in (0, 1)$, $x-5  0$, so $|x-5| = 5-x$ and $|1-x| = 1-x$.Thus, $y = (5-x) - (1-x) + x^2 = 5 - x - 1 + x + x^2 = 4 + x^2$.The area $f(1)$ is bounded by $x=0, x=1, y^2=x$ and $y=4+x^2$:$f(1) = \int_0^1 ((4+x^2) - \sqrt{x}) dx = [4x + \frac{x^3}{3} - \frac{2}{3}x^{3/2}]_0^1 = 4 + \frac{1}{3} - \frac{2}{3} = 4 - \frac{1}{3} = \frac{11}{3}$.Finally, $f(0) + f(1) = \frac{10}{3} + \frac{11}{3} = \frac{21}{3} = 7$.

Let $f(\alpha)$ denote the area of the region in the first quadrant bounded by $x=0, x=1, y^{2}=x$ and $y=|\alpha x-5|-|1-\alpha x|+\alpha x^{2}$. Then $f(0)+f(1)$ is equal to

Similar Questions

The area bounded by the curve $y=|x-2|$,$x=1$,$x=3$ and $X$-axis is

The area of the region bounded by the parabola $y = x^2 + 2$,the $X$-axis,and the lines $x = 1$ and $x = 2$ is . . . . . . sq. units.

The area bounded by the curve $y = x |\sin(\frac{x}{2})|$ and the $X$-axis between the lines $x = 0$ and $x = 4\pi$ (in square units) is: (in $\pi$)

The area of the region bounded by the curve $y = 2x - x^2$ and the $X$-axis is . . . . . . sq. units.

The area of the region bounded by the curve $y^2=4x$ and the line $y=x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module