Let's do some calculus! (23)

Calculus Level 3

$\large \displaystyle \int_{\ln \lambda}^{\ln \left( {1}/{\lambda} \right)} {\dfrac{f \left( \dfrac{x^{2}}{4} \right) \big( f(x) - f(-x) \big)}{g \left( \dfrac{x^{2}}{4} \right) \big( g(x) + g(-x) \big)}} \,dx$

Given that $f(x)$ and $g(x)$ are both continuous functions. Does the value of the above anti-derivative depend on the value of $\lambda$ ?

Notations: $\ln(\cdot)$ denotes the natural logarithm function .

For more problems on calculus, click here .

Can't say Yes No

1 solution

Chew-Seong Cheong
Sep 24, 2016

Using the identity $\displaystyle \int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx$ , we note that the integral is 0 for any $\lambda$ . $\boxed{\text{No}}$ , it is not depend on the value $\lambda$ .

$\begin{aligned} I & = \int_{\ln \lambda}^{\ln \frac 1 \lambda} \frac {f\left(\frac {x^2}4\right)(f(x)-f(-x))}{g\left(\frac {x^2}4\right)(g(x)+g(-x))} dx \\ & = \frac 12 \int_{\ln \lambda}^{\ln \frac 1 \lambda} \left( \frac {f\left(\frac {x^2}4\right)(f(x)-f(-x))}{g\left(\frac {x^2}4\right)(g(x)+g(-x))} + \frac {f\left(\frac {(\ln \frac 1\lambda + \ln \lambda - x)^2}4\right)\left(f\left(\ln \frac 1\lambda + \ln \lambda - x\right)-f\left(-(\ln \frac 1\lambda + \ln \lambda - x)\right)\right)}{g\left(\frac {(\ln \frac 1\lambda + \ln \lambda - x)^2}4\right)\left(g\left(\ln \frac 1\lambda + \ln \lambda - x\right)+g\left(-(\ln \frac 1\lambda + \ln \lambda - x)\right)\right)} \right) dx \\ & = \frac 12 \int_{\ln \lambda}^{\ln \frac 1 \lambda} \left( \frac {f\left(\frac {x^2}4\right)(f(x)-f(-x))}{g\left(\frac {x^2}4\right)(g(x)+g(-x))} + \frac {f\left(\frac {x^2}4\right)\left(f\left(- x\right)-f\left(x\right)\right)}{g\left(\frac {x^2}4\right)\left(g\left(- x\right)+g\left(x \right)\right)} \right) dx \\ & = \frac 12 \int_{\ln \lambda}^{\ln \frac 1 \lambda} 0 \ dx = 0 \end{aligned}$

Sir, in the second step you forgot to divide the expression by $2$ .

Tapas Mazumdar - 4 years, 8 months ago

Thanks. A lot of braces { } can be dropped for example, \frac 12 for $\frac 12$ , \sqrt 3 $\sqrt 3$ . 2^3 $2^3$ . Basically all behind function such as $\sin$ , $\sum$ , $\int$ do not need braces.

Chew-Seong Cheong - 4 years, 8 months ago

Won't it get confused if I have to put a long expression under them? For example if I want to show $\dfrac{12}{73}$ can I write that as \frac 12 73 ? Or $2^{n-3}$ as 2^n-3

Tapas Mazumdar - 4 years, 8 months ago

@Tapas Mazumdar – Nope, you have to write \frac{12}{73} for $\frac{12}{73}$ and 2^{n-3} for $2^{n-3}$ . They cannot be saved. But I have seen not from you where people using the template doing \int { {\sin }^{ 2 } {x} } dx instead of \int \sin^2 x dx.

You can save like \frac 1{12}, \frac {12}5 \frac \pi 2 even \int_\frac \pi 4 ^\frac \pi 2 $\int_\frac \pi 4 ^\frac \pi 2$

Chew-Seong Cheong - 4 years, 8 months ago

Perfect solution

Md Zuhair - 4 years, 4 months ago

@Chew-Seong Cheong @Md Zuhair Can we directly say that the integration of an odd continuous function is an even function and hence the expression above is always 0?

Ankit Kumar Jain - 3 years ago

If it always true?

Md Zuhair - 3 years ago

I think it is true (though not sure enough) , that is why I asked in order to confirm.

Ankit Kumar Jain - 3 years ago

@Ankit Kumar Jain – But we don't need to know that. We just need to know $\displaystyle \int_{-a}^a o(x) \ dx = 0$ and $\displaystyle \int_{-a}^a e(x) \ dx = 2 \int_0^a e(x) \ dx$ .

Chew-Seong Cheong - 3 years ago

@Chew-Seong Cheong – Sir , in your comment you mentioned that the odd integrand from $-a$ to $a$ gives 0 , isn't it the condition for the integrated function to be even?

Ankit Kumar Jain - 3 years ago

@Ankit Kumar Jain – Even function is defined as $f(x)=f(-x)$ such as $\cos(x) = \cos (-x)$ and odd function is defined as $f(x) = - f(-x)$ such as $\sin(x) = - \sin(-x)$ . They have nothing to do with the limits.

Chew-Seong Cheong - 3 years ago

@Chew-Seong Cheong – what I mean to say is that let the integration be f(x) , then as you have stated f(a) = f(-a) and is hence even.

Ankit Kumar Jain - 3 years ago

It appears that integral of odd function is an even function but I am not sure if it is always the case. Odd integrand will give 0 $\displaystyle \int_{-a}^a o(x) \ dx = 0$ instead of even integrand $\displaystyle \int_{-a}^a e(x) \ dx = 2 \int_0^a e(x) \ dx$ .

Chew-Seong Cheong - 3 years ago

Let's do some calculus! (23)

For more problems on calculus, click here .

1 solution

0 pending reports