We are asked to compare the values of three definite integrals:
The comparison depends on the integrands over the interval $[0, 1]$.
Consider the integrands $f_3(x) = e^{-x^2}$ and $f_2(x) = e^{-x^2} \cos^2x$.
For any real value of $x$, we know that $0 \le \cos^2x \le 1$.
Multiplying by $e^{-x^2}$ (which is always positive), we get: $0 \le e^{-x^2} \cos^2x \le e^{-x^2}$
Integrating this inequality from $0$ to $1$: $\int_0^1 0 \, dx \le \int_0^1 e^{-x^2} \cos^2x \, dx \le \int_0^1 e^{-x^2} \, dx$
This simplifies to $0 \le I_2 \le I_3$.
Since $\cos^2x$ is not equal to 1 over the entire interval $[0, 1]$ (it's 0 at $x = \frac{\pi}{2} + n\pi$, and less than 1 elsewhere), the inequality is strict for the integral value, meaning $I_2 < I_3$.
Consider the integrands $f_2(x) = e^{-x^2} \cos^2x$ and $f_1(x) = e^{-x} \cos^2x$.
We need to compare the terms $e^{-x^2}$ and $e^{-x}$ for $x \in [0, 1]$.
For $x$ in the interval $(0, 1)$, we have $x^2 < x$.
Multiplying by $-1$ reverses the inequality: $-x^2 > -x$.
Since the exponential function $y \mapsto e^y$ is strictly increasing, we have $e^{-x^2} > e^{-x}$ for $x \in (0, 1)$.
Given that $\cos^2x \ge 0$, we can multiply the inequality by $\cos^2x$: $e^{-x^2} \cos^2x > e^{-x} \cos^2x$ for $x \in (0, 1)$.
Integrating this inequality from $0$ to $1$: $\int_0^1 e^{-x^2} \cos^2x \, dx > \int_0^1 e^{-x} \cos^2x \, dx$
Therefore, $I_2 > I_1$.
Combining the results from the previous steps:
These inequalities together establish the order: $I_3 > I_2 > I_1$.
Let $f(x) = \int \frac{7x^{10} + 9x^8}{(1 + x^2 + 2x^9)^2} \,dx$, $x > 0$, $\lim_{x \rightarrow 0} f(x) = 0$ and $f(1) = \frac{1}{4}$.
If $A = \begin{bmatrix} 0 & 0 & 1 \\ \frac{1}{4} & f'(1) & 1 \\ \alpha^2 & 4 & 1 \end{bmatrix}$ and $B = \text{adj}(\text{adj } A)$ be such that $|B| = 81$, then $\alpha^2$ is equal to
Let $f$ be a differentiable function satisfying $f(x) = 1 - 2x + \int_{0}^{x} e^{(x - t)} f(t) dt$, $x \in \mathbf{R}$ and let $g(x) = \int_{0}^{x} (f(t) + 2)^{15} (t - 4)^{6} (t + 12)^{17} dt$, $x \in \mathbf{R}$. If $\text{p}$ and $\text{q}$ are respectively the points of local minima and local maxima of $g$, then the value of $|\text{p} + \text{q}|$ is equal to _________.
Let $f$ be a twice differentiable function such that $f(x) = \int_{0}^{x} \tan(t - x)dt - \int_{0}^{x} f(t)\tan t dt$, $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.
Then $f''\left(\frac{\pi}{6}\right) + 12f'\left(-\frac{\pi}{6}\right) + f\left(\frac{\pi}{6}\right)$ is equal to __________
If $\int \frac{2x+5}{\sqrt{7-6x-x^2}} dx$ = $A\sqrt{7-6x-x^2} + Bsin^{-1} \left( \frac{x+3}{4} \right) + C$
(Where C is a constant of integration), then the ordered pair (A,B) is equal to :-
Let $f(x) = \int \frac{7x^{10} + 9x^8}{(1 + x^2 + 2x^9)^2} \,dx$, $x > 0$, $\lim_{x \rightarrow 0} f(x) = 0$ and $f(1) = \frac{1}{4}$.
If $A = \begin{bmatrix} 0 & 0 & 1 \\ \frac{1}{4} & f'(1) & 1 \\ \alpha^2 & 4 & 1 \end{bmatrix}$ and $B = \text{adj}(\text{adj } A)$ be such that $|B| = 81$, then $\alpha^2$ is equal to