Let f(x) = int e^x (x - 1)(x - 2) dx. Then f | vedclass

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

Question

(C) Given $f(x) = \int e^x (x - 1)(x - 2) dx$.To find the interval where $f(x)$ decreases,we need to find $f'(x)$ and set $f'(x) < 0$.By the Fundament

Vedclass · Accepted Answer

$(1, 2)$

Vedclass · Answer

(C) Given $f(x) = \int e^x (x - 1)(x - 2) dx$.To find the interval where $f(x)$ decreases,we need to find $f'(x)$ and set $f'(x) By the Fundamental Theorem of Calculus,$f'(x) = \frac{d}{dx} \int e^x (x - 1)(x - 2) dx = e^x (x - 1)(x - 2)$.For $f(x)$ to be a decreasing function,we require $f'(x) So,$e^x (x - 1)(x - 2) Since $e^x > 0$ for all real $x$,the inequality simplifies to $(x - 1)(x - 2) The roots of the quadratic $(x - 1)(x - 2) = 0$ are $x = 1$ and $x = 2$.The expression $(x - 1)(x - 2)$ is negative between its roots.Therefore,$f(x)$ decreases in the interval $(1, 2)$.

Let $f(x) = \int e^x (x - 1)(x - 2) dx$. Then $f$ decreases in the interval -

Similar Questions

The value of $a$ such that $f(x) = \sqrt{3} \sin x - \cos x - 2ax + b$ decreases for all real values of $x$ is given by:

If $f(x) = \frac{\log x}{x}$ $(x > 0)$,then it is increasing in

Let $f(x)=e^x-x$ and $g(x)=x^2-x, \forall x \in R$. Then the set of all $x \in R$,where the function $h(x)=(fog)(x)$ is increasing is

Prove that the function $f$ given by $f(x) = \log(\sin x)$ is increasing on $\left(0, \frac{\pi}{2}\right)$ and decreasing on $\left(\frac{\pi}{2}, \pi\right)$.

$A$ function is matched below against an interval where it is supposed to be increasing. Which of the following pairs is incorrectly matched?
Interval | Function

Vedclass Test Series

Exam Paper Generator

Online Exam Module