If [t] denotes the greatest integer leq t,the | vedclass

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

Question

(A) Let $f(x) = 2x - |3x^2 - 5x + 2| + 1$. The expression inside the greatest integer function is $g(x) = 2x - |(3x-2)(x-1)| + 1$.For $x \in [0, 2/3]$

Vedclass · Accepted Answer

$\frac{\sqrt{37}+\sqrt{13}-4}{6}$

Vedclass · Answer

(A) Let $f(x) = 2x - |3x^2 - 5x + 2| + 1$. The expression inside the greatest integer function is $g(x) = 2x - |(3x-2)(x-1)| + 1$.For $x \in [0, 2/3]$,$3x^2 - 5x + 2 \geq 0$,so $|3x^2 - 5x + 2| = 3x^2 - 5x + 2$. Then $g(x) = 2x - (3x^2 - 5x + 2) + 1 = -3x^2 + 7x - 1$.For $x \in [2/3, 1]$,$3x^2 - 5x + 2 \leq 0$,so $|3x^2 - 5x + 2| = -(3x^2 - 5x + 2)$. Then $g(x) = 2x + (3x^2 - 5x + 2) + 1 = 3x^2 - 3x + 3$.Evaluating the integral $\int_{0}^{1} [g(x)] dx$ involves splitting the interval based on the integer values of $g(x)$.For $x \in [0, 2/3]$,$g(x) = -3x^2 + 7x - 1$. The roots of $g(x) = k$ are found using the quadratic formula.After calculating the integral over the sub-intervals where $[g(x)]$ is constant,we obtain the result:$I = \frac{\sqrt{37} + \sqrt{13} - 4}{6}$.

If $[t]$ denotes the greatest integer $\leq t$,then the value of $\int_{0}^{1}\left[2 x-\left|3 x^{2}-5 x+2\right|+1\right] d x$ is.

Similar Questions

$\int_0^{2\pi } {\sqrt {1 + \sin \frac{x}{2}} \,dx = } $

$\int_0^\infty {{e^{ - 2x}}(\sin 2x + \cos 2x)\,dx = } $

The value of the integral $\int_{\pi/6}^{\pi/3} \frac{4 - \csc^2 x}{\cos^4 x} dx$ is:

Evaluate the definite integral $\int_0^3 [x] \, dx$,where $[x]$ denotes the greatest integer function.

If $[x]$ is the greatest integer function not greater than $x$,then $\int_0^8 [x] dx$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module