The value 9 int_0^9 left[ sqrt{frac{10x}{x+1} | vedclass

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

Question

(A) Let $f(x) = \sqrt{\frac{10x}{x+1}} = \sqrt{\frac{10(x+1)-10}{x+1}} = \sqrt{10 - \frac{10}{x+1}}$.As $x$ increases from $0$ to $9$,$f(x)$ increases

Vedclass · Accepted Answer

$155$

Vedclass · Answer

(A) Let $f(x) = \sqrt{\frac{10x}{x+1}} = \sqrt{\frac{10(x+1)-10}{x+1}} = \sqrt{10 - \frac{10}{x+1}}$.As $x$ increases from $0$ to $9$,$f(x)$ increases from $0$ to $3$.The value of $[f(x)]$ changes at values where $f(x) = k$ for $k \in \{1, 2, 3\}$.For $f(x) = 1$: $\frac{10x}{x+1} = 1 \Rightarrow 10x = x+1 \Rightarrow x = \frac{1}{9}$.For $f(x) = 2$: $\frac{10x}{x+1} = 4 \Rightarrow 10x = 4x+4 \Rightarrow 6x = 4 \Rightarrow x = \frac{2}{3}$.For $f(x) = 3$: $\frac{10x}{x+1} = 9 \Rightarrow 10x = 9x+9 \Rightarrow x = 9$.Thus,the integral $I = 9 \int_0^9 [f(x)] dx$ becomes:$I = 9 \left( \int_0^{1/9} 0 dx + \int_{1/9}^{2/3} 1 dx + \int_{2/3}^9 2 dx \right)$.$I = 9 \left( 0 + (\frac{2}{3} - \frac{1}{9}) + 2(9 - \frac{2}{3}) \right)$.$I = 9 \left( \frac{5}{9} + 2(\frac{25}{3}) \right) = 9 \left( \frac{5}{9} + \frac{50}{3} \right) = 5 + 150 = 155$.

The value $9 \int_0^9 \left[ \sqrt{\frac{10x}{x+1}} \right] dx$,where $[t]$ denotes the greatest integer less than or equal to $t$,is . . . . . . .

Similar Questions

The value of the integral $\int_{-\pi}^{\pi} (\cos ax - \sin bx)^2 dx$,where $a$ and $b$ are integers,is

If $\int_0^{\frac{\pi}{3}} \cos^4 x \, dx = a\pi + b\sqrt{3}$,where $a$ and $b$ are rational numbers,then $9a + 8b$ is equal to:

The approximate value of $\int_{1}^{9} x^2 dx$ by using the trapezoidal rule with $4$ equal intervals is:

If $\int_{0}^{1}(5x^{2}-3x+k)dx=0$,then $k=$

$\int_0^{\frac{\pi}{2}} x^3 \sin x \, dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module