Can you solve this integral?

Thanks Snehashish for the suggestion! This is from the JEE Mains 2020 test in India.

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[sp_od]

You don't even need the outer absolute value because on the interval -pi to pi, the inner function is already always non-negative

[MikedeKokkie]

If you're going to go according to the graph anyway, you can see that the two triangles combine into a square of length pi, so of course the answer is then pi squared, yes...

[Shreyas_Jaiswal]

I did like this.
First draw |x| as V, then draw -|x| as ^, then push the entire curve upwards by π, so you get a triangle between -π and +π with height π, this gives the area =1/2×2π×π=π².

[maxwellwright]

The function to integrate is even, so it is equal to 2 times the integral from 0 to π of the function.
When x is between 0 and π, π-x>0. So the function to integrate is π-x. We get π²

[nvapisces]

Kinda just imagined the whole graph and did the calculation in my head lol. You just need to know how graph transformations work

[vitalsbat]

Even function so we can do 2x int from 0 to pi
Then since x is between 0 to pi, outer absolute value is useless
Then it's simple

[jobogella]

just a small thing
under these circumstances pi - x can be equal to 0

[tryingtomakeanamebelike]

The graph is certainly neat, but you can just solve this normally

[geoffreymak]

It’s really easy… just visualize, no arithmetic

[DrDeuteron]

the function is clearly even, so 2 x [0, pi] integral is just fine, . and 1 side is pi^2 - pi^2/2

[racheline_nya]

pi-|x| is greater than or equal to 0, but not necessarily greater than 0
(it is equal to 0 at the endpoints)

[FatihKarakurt]

Notice that it's an even function; half the range and get rid of all absolute values.

[matthewpowers]

Pi squared obviously, just think of it graphically

[TheEulerID]

You don't even need to use calculus to calculate this one, although it works just as well. Just plot it out and you observe two back-to-back right angle isosceles triangles of length pi in the x & y axis. The area must be pi^2.

There really isn't anything tricky about this one once you plot it out.

[laurendoe]

While this is all true, it really doesn't invoke integration.

[renaultpontiac]

No tricky at all suppose f(x) =pi- abs(x)
f is defined in R so for every x in R (-x) is in R also
f(-x) =f(x) so f is pair
Which mean to integrate from -pi to pi you just evaluate the double integral from 0 to pi of pi-x
Antiderivative of pi-x is xpi-1/2 x^2 so the integral is2[ pi^2-1/2pi^2]=2*1/2 pi^2=pi^2

[zizo-veib]

I personally approached it in a different way without graphs or cases, from the definition of absolute it's an even function meaning I can take out the absolute and write it as 2x the integral of same from 0 to π, distribute the integral on π and |x| the integral of
|x| is from 0 to π so no need for the absolute at all just write it as x and integral solved

[malicacidissour]

Shouldn’t it be Pi + x > or = 0 and Pi - x > or = 0, since x is > or = Pi and in the second it’s > or = - Pi, so it could be Pi - Pi which is 0.

[Lucky-swbv]

This is the exact same question that we have in our class 12th in maths book.
😂

[chandrashekharmehta]

Que. From jee mains exam India..
And it is not tricky
simply draw tha graph and get answer