Integral Comparison Analysis
We are asked to compare the values of three definite integrals:
- $I_1 = \int_0^1 e^{-x} \cos^2x \, dx$
- $I_2 = \int_0^1 e^{-x^2} \cos^2x \, dx$
- $I_3 = \int_0^1 e^{-x^2} \, dx$
The comparison depends on the integrands over the interval $[0, 1]$.
Comparing $I_3$ and $I_2$
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 $I_2 < I_3$.
Comparing $I_2$ and $I_1$
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, $I_2 > I_1$.
Final Integral Ordering
Combining the results from the previous steps:
- $I_2 < I_3$
- $I_2 > I_1$
These inequalities together establish the order: $I_3 > I_2 > I_1$.
Similar Questions
Let [.] denote the greatest integer function. If $\int_{0}^{e^3} \left[\frac{1}{e^{x-1}}\right] dx = \alpha - \log_e 2$, then $\alpha^3$ is equal to ____________.
Let $f: R\to R$ be a thrice differentiable odd function satisfying $f'(x)\ge0, f''(x)=f(x), f(0)=0, f'(0)=3$. Then $9f(\log_e 3)$ is equal to ___________.
- If $\int \frac{(\sqrt{1+x^2}+x)^{10}}{(\sqrt{1+x^2}-x)^9} dx = \frac{1}{m} \left( (\sqrt{1+x^2}+x)^n (n\sqrt{1+x^2}-x) \right) + C$ where $C$ is the constant of integration and $m, n \in N$, then $m+n$ is equal to
- Let $y = y (x)$ be the solution curve of the differentialequation $x (x^2 + e^x) dy + (e^x (x-2) y-x^3) dx = 0, x > 0$, passing through the point $(1, 0)$.Then $y (2)$ is equal to
The integral $\int_{-1}^{2} (\pi^2 x \sin (\pi x))dx$ is equal to :
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(Where C is a constant of integration), then the ordered pair (A,B) is equal to :-
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- Let $(a, b)$ be the point of intersection of the curve $x^2 = 2y$ and the straight line $y -2x-6=0$ in the second quadrant. Then the integral $I = \int_{a}^{b} \frac{9x^2}{1 + 5^x} dx$ is equal to :
- Let $f: [1, \infty) \to [2, \infty)$ be a differentiable function. If $10 \int_{1}^{x} f(t)dt = 5xf(x) - x^5 - 9$ for all $x\ge1$, then the value of $f(3)$ is :
- Let $[\cdot]$ denote the greatest integer function and $f(x) = \lim_{n \to \infty} \frac{1}{n^3} \sum_{k=1}^n \left[ \frac{k^2}{3^x} \right]$. Then $12 \sum_{j=1}^\infty f(j)$ is equal to __________.
Important Questions from Integral Calculus
Let [.] denote the greatest integer function. If $\int_{0}^{e^3} \left[\frac{1}{e^{x-1}}\right] dx = \alpha - \log_e 2$, then $\alpha^3$ is equal to ____________.
Let $f: R\to R$ be a thrice differentiable odd function satisfying $f'(x)\ge0, f''(x)=f(x), f(0)=0, f'(0)=3$. Then $9f(\log_e 3)$ is equal to ___________.
- If $\int \frac{(\sqrt{1+x^2}+x)^{10}}{(\sqrt{1+x^2}-x)^9} dx = \frac{1}{m} \left( (\sqrt{1+x^2}+x)^n (n\sqrt{1+x^2}-x) \right) + C$ where $C$ is the constant of integration and $m, n \in N$, then $m+n$ is equal to
- Let $y = y (x)$ be the solution curve of the differentialequation $x (x^2 + e^x) dy + (e^x (x-2) y-x^3) dx = 0, x > 0$, passing through the point $(1, 0)$.Then $y (2)$ is equal to
The integral $\int_{-1}^{2} (\pi^2 x \sin (\pi x))dx$ is equal to :
