Question
Let $f(x)=\int e^{x}(x-1)(x-2) d x$, then $f$ decreases in the interval(a) $(-\infty,-2)$(b) $(-2,-1)$(c) $(1,2)$(d) $(2, \infty)$
Step 1
By the Leibniz rule (also known as the Fundamental Theorem of Calculus), the derivative of an integral with a variable upper limit is simply the integrand evaluated at the upper limit. Therefore, we have \[f'(x) = e^{x}(x-1)(x-2).\] Show more…
Show all steps
Your feedback will help us improve your experience
Aman Gupta and 95 other Calculus 1 / AB educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
The function $f(x)=\frac{x}{4+x^{2}}$ decreases on the interval (a) $(-\infty,-1)$ (b) $(-\infty, 0)$ (c) $(-\infty,-2) \cup(2, \infty)$ (d) $(-2,2)$
Monotonocity
Level II
If $f(x)=\int_{x^{2}}^{x^{2}+1} e^{r^{2}} d t$, then the interval in which $f(x)$ is inc. is (a) $(0, \infty)$ (b) $(-\infty, 0)$ (c) $[-2,2]$ (d) no where
The function $f(x)=|x+2|+|x-1|$ is (A) increasing in $(1, \infty)$ (B) increasing in $[1, \infty)$ (C) decreasing in (-\infty, - 2] (D) decreasing in $(-\infty,-2)$
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD