Let f(x) = intlimits_0^x {{e^{ - {t^2}}}dt} | vedclass

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

Question

(D) Consider the function $g(t) = e^{-t^2}$. For $t \in (0, x)$,the function $g(t)$ is strictly decreasing because $g'(t) = -2t e^{-t^2}

Vedclass · Accepted Answer

$f(x) > \frac{x}{2}\left( {1 + {e^{ - {x^2}}}} \right)$

Vedclass · Answer

(D) Consider the function $g(t) = e^{-t^2}$. For $t \in (0, x)$,the function $g(t)$ is strictly decreasing because $g'(t) = -2t e^{-t^2}  0$.Since $g(t)$ is a strictly decreasing concave function on $(0, x)$,the area under the curve $y = e^{-t^2}$ from $0$ to $x$ is greater than the area of the trapezoid formed by the points $(0, 0), (x, 0), (x, e^{-x^2}),$ and $(0, 1)$.The area of this trapezoid is given by $\frac{1}{2} 	imes (	ext{sum of parallel sides}) 	imes (	ext{height}) = \frac{1}{2} (1 + e^{-x^2}) 	imes x$.Therefore,$f(x) = \int_0^x e^{-t^2} dt > \frac{x}{2} (1 + e^{-x^2})$.

Let $f(x) = \int\limits_0^x {{e^{ - {t^2}}}dt} $ for all $x > 0$. Then for all $x > 0$:

Similar Questions

If $\int_0^{k} \frac{d x}{2+8 x^2}=\frac{\pi}{16}$,then the value of $k$ is

$f(x) = \begin{cases} x^2, & 0 \leq x < 1 \\ \sqrt{x}, & 1 \leq x \leq 2 \end{cases} \implies \int_0^2 f(x) \, dx = ?$

The maximum value of the function $f(x) = \int\limits_0^1 {t\,\sin \left( {x + \pi t} \right)} dt$ for $x \in R$ is

The value of $\int_{0}^{1} \left( \prod_{r=1}^{n} (x+r) \right) \left( \sum_{k=1}^{n} \frac{1}{x+k} \right) dx$ equals

$\int_{\frac{\pi}{4}}^{\frac{\pi}{4}} \log_e(\sin x + \cos x) \, dx$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module