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+18 votes

Let $S = \sum_{i=3}^{100} i \log_{2} i$, and $T = \int_{2}^{100} x \log_{2}x dx$.

Which of the following statements is true?

  1. $S > T$
  2. $S = T$
  3. $S < T$ and $2S > T$
  4. $2S ≤ T$
asked in Calculus by Veteran (52k points)
edited by | 1.7k views

3 Answers

+16 votes
Best answer

$x\log_2 x$ is a continuously increasing function, and for a continuously increasing function $f(x)$,

$$\sum_{x=a}^{b} f(x) > \int_a^b f(x)dx$$

But in question, summation of L.H.S. above, $a=3$ and in R.H.S, $a=2$, so we don't know whether $S > T$. So we compute some initial values :

$\sum_{x=3}^{4} x\log_2 x \approx 12.754$, and $\int_2^4 x\log_2 x = 11$

Since $\sum_{x=3}^{4} x\log_2 x > \int_2^4 x\log_2 x$, and since we already know that

$\sum_{x=5}^{100} x\log_2 x > \int_5^{100} x\log_2 x$

So $\sum_{x=3}^{100} x\log_2 x > \int_2^{100} x\log_2 x$

So S > T, and option (A) is correct.

answered by Boss (11.4k points)
selected by
Continuously increasing function rt?

And what is the reason for this inequality?

$\sum_{x=a}^{b} f(x) > \int_a^b f(x)dx$

Yes, continuously increasing. I though of it intuitively, if a function is increasing, then all the rectangles (for approximating the area under the curve) will lie below the function, and hence the integral evaluated will be less than actual area under the curve.

That's nice explanation :)

But how $\sum{f(x)}$ is comparable to $\int{f(x) dx}$ ?

the latter gives us the area under the curve it has different units(see $dx$ exists in the expression too). while the former provides the sum of all values of $f(x)$ as per the definition given.

That could be a reason why escaped from putting $dx$ while providing the solution.

I guess only numerical values are to be compared. 

also, Answer given is more of an intuitive solution; Not a proof why for a continuous and increasing function $f(x)$ : $$\sum_{x=a}^{b} f(x) > \int_a^b f(x)dx$$

here, $f(x) = x \log_2(x)$:

@Happy_Mittal, Sir according to your explaination, the value of integration should be greater than summation because integration finds sum for infinitesimal intervals, so it better approximates the sum under curve than summation. Please clarify it.
The first assumption is wrong if the function is increasing then the integral is larger than summation , please clearify on the this point.

In then graph posted by @Happy Mittal ... the integral evaluated will be less than actual area under the curve.

 Here area under curve is same as integral of function right ??

And sum of area of all rectangle is equivalent to summation right ??

Any one plzz clear my doubt ...  Acc. to me integral should be greater than summation.

Happy Mittal Sir, can you please explain a bit more the reason for which the summation is greater than integral ?

that's why they give i from 3 , not from 2.


if we see the value of "S" is quite large

The rectangles do not represent summation, summation as given in the question is not area under the curve, just sum of the function at different values of x. Also you have assumed left Riemann sum, why not assume right Reimann sum? The entire conclusion will be opposite if right Reimann sum is used.

@Happy Mittal How to do the calculation in exam without calculator

For continuously increasing functions, it's the other way round right ?

$\int_{a}^{b} f(x).dx > \sum_{a}^{b} f(i)$

If you take left Riemann sum, that is. There's some portion that's not covered by the rectangle, but in the integral, it would be included.

Reference from Cormen Appendix $A$.

Since function $xlog_{2}x$ is monotonically increasing from point $x=1$ because at $x=1$ , it is zero because $log\;1 = 0$. 

So,  According to given relation in above image in case of monotonically increasing function, we can write it as :-

$\int_{m-1}^{n}f(x)\;dx \leqslant \sum_{k=m}^{n} f(k)$

So, For the given , we can write the relation between sigma and integral in the given limit as :-

$\int_{2}^{100}xlog_{2}x\;dx \leqslant \sum_{i=3}^{100} ilog_{2}i$

thanks prateek. this idea helps in solving the problem faster.

@Happy Mittal I think your intuition works only for left Riemann sums(only top left corners of rectangles touch the curve). If we use a right Riemann sum (only top right corners of rectangles touch the curve), each rectangle will have some area which is above the curve (only for continuously increasing functions). Thus the sum of the areas of rectangles in right Riemann sum will be greater than area under the curve. Please correct me if I am wrong. Link which explains left and right Riemann sums

0 votes



i.e. S<T


if 2S>T


S>2  (that is true )

so 2S>T

and answer is C

answered by (205 points)
You assumed integral is same as summation which is not always true.

Also, your argument for second one is wrong. You assumed "2S > T" and went to get "S > 2" which is true and hence took "2S > T" as true. This is a wrong proof method. If you take an assumption, you should derive a "CONTRADICTION" and then you can say "ASSUMPTION" is false. You cannot derive a "TRUE" from assumption and then say assumption is TRUE.
Also, here is the proof that S > T

The difference (S -T) there is approx 229.81
0 votes

answer = option A

$x \log_2(x)$ is a continuously increasing function for the interval [2, 100]

We need to compare the numerical values 
S=$\sum_{i=3}^{100} i \log_{2} i$
T=$\int_{2}^{100} x \log_{2}x dx$

$\sum_{x=3}^{4} x\log_2 x \approx 12.754$ and
$\int_2^4 x\log_2 x\ dx= 11$

we compute the actual values as:



where it is seen that S > T

answered by Boss (30.6k points)
edited by
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