Practicing Success
The value of the integral $\int\limits_0^1 e^{x^2} d x$ lies in the interval |
(0, 1) (-1, 0) (1, e) none of these |
(1, e) |
Since $e^{x^2}$ is an increasing function on $(0,1)$. ∴ $m=e^0=1$ and $M=e^1=e$, where $m$ and $M$ are minimum and maximum values of $f(x)=e^{x^2}$ in the interval $(0,1)$ for all $x \in(0,1)$ $\Rightarrow 1(1-0)<\int\limits_0^1 e^{x^2} d x<e(1-0) \Rightarrow 1<\int\limits_0^1 e^{x^2} d x<e$ |