Area of the region (in sq. units) bounded by | vedclass

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

Question

(B) The area $A$ is given by the integral $\int_{0}^{2} |y| dx = \int_{0}^{2} |x^2-5x+4| dx$. First,find the roots of $x^2-5x+4=0$,which are $(x-1)(x-

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) The area $A$ is given by the integral $\int_{0}^{2} |y| dx = \int_{0}^{2} |x^2-5x+4| dx$. First,find the roots of $x^2-5x+4=0$,which are $(x-1)(x-4)=0$,so $x=1$ and $x=4$. In the interval $[0, 1]$,$x^2-5x+4 \geq 0$. In the interval $[1, 2]$,$x^2-5x+4 \leq 0$. Thus,$A = \int_{0}^{1} (x^2-5x+4) dx + \int_{1}^{2} -(x^2-5x+4) dx$. Evaluating the first integral: $[\frac{x^3}{3} - \frac{5x^2}{2} + 4x]_{0}^{1} = \frac{1}{3} - \frac{5}{2} + 4 = \frac{2-15+24}{6} = \frac{11}{6}$. Evaluating the second integral: $-[\frac{x^3}{3} - \frac{5x^2}{2} + 4x]_{1}^{2} = -[(\frac{8}{3} - 10 + 8) - (\frac{1}{3} - \frac{5}{2} + 4)] = -[(\frac{2}{3}) - (\frac{11}{6})] = -[\frac{4-11}{6}] = -[-\frac{7}{6}] = \frac{7}{6}$. Total Area $A = \frac{11}{6} + \frac{7}{6} = \frac{18}{6} = 3$ sq. units.

Area of the region (in sq. units) bounded by the curve $y=x^2-5x+4$,$x=0$,$x=2$ and the $X$-axis is

Similar Questions

The area of the region bounded by $y = |x - 1|$ and $y = 1$ is

The area bounded by the $y-$axis,$y=\cos x$ and $y=\sin x$ when $0 \leq x \leq \frac{\pi}{2}$ is

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

Using integration,find the area of the triangular region whose sides have the equations $y=2x+1$,$y=3x+1$,and $x=4$.

Let $A$ denote the area bounded by the curve $y=\frac{1}{x}$ and the lines $y=0, x=1, x=10$. Let $B=1+\frac{1}{2}+\ldots+\frac{1}{9}$ and $C=\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{10}$. Then,

Vedclass Test Series

Exam Paper Generator

Online Exam Module