Question

Another very interesting and challenging integration problem

Answer 1

If \(\large{f(x) = [\cfrac{x}{120} + \cfrac{x^3}{30}]}\) defined in [0,3] then

\[\large{\int\limits_{f(x)}^{f(x) + 1} (f(x) + 2)dx} = ?\]

where([.] denotes greatest integer function)

Options are:

(a) 1
(b) 2
(c) 0
(d) None of these

Answer 2

ok, so what is the value of f(0)

Answer 3

Greatest integer function means the greatest integer less than or equal to real number x.

Answer 4

f(0) = 0

Answer 5

I am aware

Answer 6

Now, f(3)+1?

Answer 7

Gonna go with (b)

Answer 8

Good

Answer 9

Your reason ?

Answer 10

\(f(x)=0\) for \(x\in[0,3]\), so all the \(f(x)\) can be replaced with 0, giving you
\[\int_0^12~dx=2(1-0)=2\]

Answer 11

Great work @SithsAndGiggles.

For those who don't know how @SithsAndGiggles got f(x) = 0

You can verify by differentiating x/120 + x^3/30 that this is an increasing function.

So, minimum value of this will be 0 (by putting x=0) and maximum will be 37/40 (by putting x = 3). So,

f(x) = 0

Answer 12

This was another such type of problem where you must not judge the question by just looking it.

Answer 13

Do you all want to solve some more such questions ?

Answer 14

Yes please!

Answer 15

Determine the largest number in the infinite sequence :

\[1 , 2^{\cfrac{1}{2}}, 3^{\cfrac{1}{3}}, ... , n^{\cfrac{1}{n}},...\]

Answer 16

3^1/3 ...

Answer 17

correct but why ?

Answer 18

after 3^1/3 ... the sequence is decreasing.

till 3^1/3 it increases and then decreases

Answer 19

This involves a very important result in solution.

A good method @mathslover

Answer 20

But can you prove this ?

Answer 21

Square each term
1 .. 2 .. 3^(2/3) .. 2.... n^(2/n)....
Cube each term
1 ... 8 ... 9 ...8.... n^(6/n) ...


if n > 6 then n^(6/n) < n

if n < 6 then n < n^(6/n)

Answer 22

I'm trying to prove it... this is a new method :-)

Answer 23

you can use calculus

Answer 24

Or binomial theorem

Answer 25

It is clear that, 9^n > n^6

right?

Answer 26

for n > 1

Answer 27

And how is it clear ?

You have to justify your each and every step.

Answer 28

Its very easy to solve this by calculus

Answer 29

Why not to prove that by induction

Answer 30

First case =>

9 > 8

Further case

(assuming 3^n > n^3)

3^(n + 1) > (n+1)^3

3^n * 3 > 3n^3

Answer 31

\(3n^3 \ge (n+1)^3\)

Therefore : \(3^n \times 3 \ge (n+1)^3 \)

For n > 1

then :

\(3^n \times 3 > (n+1)^3\)

Thus :

\(3^n > n^3\)

Answer 32

Thus :

9^n > n^6

And we had :

1 , 8 , 9^n , ... , n^(6) , ....

AS proved above : 9^n > n^6

Thus, we can say, 9^n will be the greatest term.

OR : 3^(1/3) will be the greatest term in the original sequence.

Answer 33

Great answer @mathslover

Answer 34

Thanks a lot. I'm now looking at your calculus method.

Answer 35

Let:

\(\large{y = x^{\cfrac{1}{x}}}\)

Now take log both sides:

\(\large{\ln{y} = {\cfrac{1}{x}}\ln{x}}\)

Differentiate both sides,

\(\large{\cfrac{1}{y}{\cfrac{dy}{dx}} = {\cfrac{1}{x^2}} - {\cfrac{\ln{x}}{x^2}}}\)

\(\large{\implies{\cfrac{dy}{dx}} = {x^{\cfrac{1}{x}}}[{{\cfrac{1}{x^2}} - {\cfrac{\ln{x}}{x^2}}}]}\)

Now put \(\cfrac{dy}{dx} = 0\).

So, I get, \(\large{\ln{x} = 1}\), or, x = e.

Now, by first principle, I can check that x = e is point of maxima.

Now, since \(\large{3^{\cfrac{1}{3}} > 2^{\cfrac{1}{2}}}\). So,

maximum value will be \(3^{\cfrac{1}{3}}\).