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0 answers
45 views

Prove/disprove upper bound and lower bound of the Integral

Hey I need to Prove or disprove this sentence: $$ \frac{4}{9}(e-1) \leq \int_0^1 \frac{e^x}{(1+x)(2-x)} \, dx \leq \frac{1}{2}(e-1) $$ using the infimum and supremum method for integrals, where m and ...
miiky123's user avatar
0 votes
1 answer
26 views

Double integral: technique to derive the limitations of $y$ (or $x$)

$$\iint_Cy dxdy, \quad C=\{(x,y)\colon0\leqslant x\leqslant2+y-y^2\}.$$ It is simple to see $x=2+y-y^2$ is a parabola with the symmetry axes is $x$ and the vertex $(9/4,1/2)$. It is easy to find the ...
Sebastiano's user avatar
  • 7,792
2 votes
0 answers
206 views

Control $\int_0^\infty |\psi(x)|^2 dx$ by $\int_0^\infty \int_0^\infty K(x+y)\psi(x) \psi(y) dxdy$

Assume that $\psi(x)$ is bounded and integrable on $x \in [0,\infty)$ with $\int_0^\infty \psi(x) dx = 0$, and suppose that $K \colon (0,\infty) \to (0,\infty)$ is some kernel function satisfying $K(x)...
Fei Cao's user avatar
  • 2,860
0 votes
1 answer
73 views

Prove that $\int_0^1 e^{-tu}(1-u)^{\alpha}du\leq t^{-1}$ for $\alpha,t>0$

Let $\alpha>0$, I need to prove that there exists $t_0>0$ such that $$\int_0^1 e^{-tu}(1-u)^{\alpha}du\leq t^{-1}, \forall t>t_0.$$ I received help and found that by Watson's Lemma you could ...
Rodrigo M's user avatar
1 vote
1 answer
1k views

Determining Bounds of Theta for Double Integration of a Polar Region

Given the region $$ R:(x, y | x^2 + y^2 \le 4x) $$ And given the function $f(x, y) = \frac{x}{\sqrt{x^2+y^2}}$, find the double integral of the polar region. So upon sketching the graph we get a ...
Stephen's user avatar
  • 11
0 votes
1 answer
255 views

Convolution of function... integral bounds?

Okay so for this question: Convolution of a function with itself The answer stated that in the case of $x\le 0$: the integral bounds are from 0 to x. Why is this? I also don't understand why from $...
boolprop's user avatar
2 votes
2 answers
65 views

Lower bound $\int_0^\infty e^{-t-\frac{t^2}{2\sigma^2}}dt$ by $1-\frac{1}{\sigma^2}$

I am trying to show a lower bound $\int_0^\infty e^{-t-\frac{t^2}{2\sigma^2}}dt \geq 1-\frac{1}{\sigma^2}$. It seems like one could try integration by parts and get $$ \int_0^\infty e^{-t-\frac{t^2}{2\...
HOIJkj's user avatar
  • 97
0 votes
2 answers
183 views

Double Integral Bounds / Substitution

I am having trouble following these steps in a reading on multivariable calculus. Due to a change of variables: $ \displaystyle\int_0^1 \int_0^s v^7 dv \, ds = \...
Zak Fischer's user avatar
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0 answers
30 views

Multivariable Calculus - Change of Bound Help!

I am having trouble following these steps in a reading on multivariable calculus. Due to a change of variables: $ \displaystyle\int_t^T \int_t^s \theta_v dv \, ...
Zak Fischer's user avatar