Well, I was doing some problems where I could use the technique mentioned above. But I was so much confused about the constant we have to add at the end. In most of the problems, the constant was turning out to be the answer so I couldn't work out a guess even. Can anyone please help? I checked out Wikipedia and they say "By definition \(I(\text{something}) = \text{something else}\)" -_- How does one know the definition of integral? Kindly enlighten me!

## Comments

Sort by:

TopNewestCan you please state your question more clearly? Maybe you can tell us one of the problems which confused you?

I hope the following is of some help:

Consider a known function \(f(x)\). I think you may be familiar with the concept of derivative w.r.t \(x\) of \(f(x)\). This is simply the rate of change of \(f\) w.r.t \(x\) for some \(x\).

Now, on the same lines... why don't we think backwards? Given \(f(x)\), can we find a function \(g(x)\) such that the derivative of \(g(x)\) w.r.t \(x\) is same as \(f(x)\)? The answer to this question is yes, and it is possible through integration. This means:

\(\frac { d }{ dx } (g\left( x \right) )=f\left( x \right) \) or:

\[g\left( x \right) =\int { f\left( x \right) dx } \]

There are different methods that we can use to evaluate such integrals. But here, we will stick to discussing the significance of the constant.

If we observe, the answer to our question is satisfied by infinitely many \(g(x)\), each of which differ from one another by a constant. This is due to the fact that while differentiating \(g(x)\), the constant term vanishes and hence, to get the whole set of functions which on differentiating yield \(f(x)\), we add a constant to the integral after integrating. – Raghav Vaidyanathan · 2 years, 4 months ago

Log in to reply

this! I must have shared the link previously only. When we integrate \(I'(\alpha)\), we have to add a constant of integration. Now, my question is how can we get that constant. – Kartik Sharma · 2 years, 4 months ago

You totally misunderstood my question. "Differentiation under integral sign" is a technique used to integrate various integrals made famous particularly by Feynman. Check outLog in to reply

You have f(y) +C= Integral of g(x) with respect to x from x=0 to x=y.

It is easy to observe that f(0)=0

With this initial value you can find the constant . – Ronak Agarwal · 2 years, 4 months ago

Log in to reply

f(y) + C = integral of g(x,y) w.r.t x from x =0 to x = a

for any constant a.

Then, what can we do? – Kartik Sharma · 2 years, 4 months ago

Log in to reply

Eg, g(x,y)= e^(-y).sin(x) , put y= infinity. – Ronak Agarwal · 2 years, 4 months ago

Log in to reply

– Kartik Sharma · 2 years, 4 months ago

Hmm, okay I will try! Thanks!Log in to reply

by definition, what does that mean? – Kartik Sharma · 2 years, 4 months agoLog in to reply

– Raghav Vaidyanathan · 2 years, 4 months ago

They say that it is zero by definition because the integrand becomes zero everywhere if we put \(\alpha= \pi/2\).Log in to reply

– Kartik Sharma · 2 years, 4 months ago

Hmm okay now I see! Thanks! BTW, I was doing your problems only when I got this doubt!Log in to reply

@Azhaghu Roopesh M @Ronak Agarwal @Raghav Vaidyanathan @Brian Charlesworth @Pratik Shastri @Jon Haussmann and more! – Kartik Sharma · 2 years, 4 months ago

Log in to reply