[x] represents the greatest integer function. | vedclass

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

Question

(D) We evaluate the integral $\int_{\sqrt{3}}^{\sqrt{18}}[x] \, dx$ by splitting the interval based on the values where $[x]$ changes: $\int_{\sqrt{3}

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(D) We evaluate the integral $\int_{\sqrt{3}}^{\sqrt{18}}[x] \, dx$ by splitting the interval based on the values where $[x]$ changes: $\int_{\sqrt{3}}^{\sqrt{18}}[x] \, dx = \int_{\sqrt{3}}^{2}[x] \, dx + \int_{2}^{3}[x] \, dx + \int_{3}^{4}[x] \, dx + \int_{4}^{\sqrt{18}}[x] \, dx$ Since $\sqrt{3} \approx 1.732$ and $\sqrt{18} = 3\sqrt{2} \approx 4.242$,we have: $\int_{\sqrt{3}}^{2} 1 \, dx + \int_{2}^{3} 2 \, dx + \int_{3}^{4} 3 \, dx + \int_{4}^{3\sqrt{2}} 4 \, dx$ $= (2 - \sqrt{3}) + 2(3 - 2) + 3(4 - 3) + 4(3\sqrt{2} - 4)$ $= 2 - \sqrt{3} + 2 + 3 + 12\sqrt{2} - 16$ $= (2 + 2 + 3 - 16) + 12\sqrt{2} - \sqrt{3}$ $= -9 + 12\sqrt{2} - \sqrt{3}$ Comparing this with $a + b\sqrt{2} + c\sqrt{3}$,we get $a = -9$,$b = 12$,and $c = -1$. Therefore,$a + b + c = -9 + 12 - 1 = 2$.

$[x]$ represents the greatest integer function. If $\int_{\sqrt{3}}^{\sqrt{18}}[x] \, dx = a + b\sqrt{2} + c\sqrt{3}$,then $a + b + c =$

Similar Questions

$\int_0^2 \frac{x^3}{(x^2 + 1)^{3/2}} \, dx = $

$ \int_{-5}^{5} |x+2| \, dx $ is equal to

If $[x]$ denotes the greatest integer function,then $\int_0^5 x^2[x] d x=$

$\int_0^4 |2x - 5| \, dx = $

If $f(x) = \begin{cases} \sqrt{1-x} & 0 \leqslant x \leqslant 1 \\ (7x-6)^{-1/3} & 1 < x \leqslant 2 \end{cases}$,then $\int_{0}^{2} f(x) dx$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module