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?

- $S > T$
- $S = T$
- $S < T$ and $2S > T$
- $2S ≤ T$

### 1 comment

## 3 Answers

$x\log_2 x$ 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.**

### 18 Comments

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

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.

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 :-

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

@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 https://www.khanacademy.org/math/ap-calculus-ab/ab-integration-new/ab-6-2/a/left-and-right-riemann-sums

### 2 Comments

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.

http://www.wolframalpha.com/input/?i=%28sum+xlogx+from+x%3D3+to+100%29+-+%28integrate+xlogx+from+x%3D2+to+100%29+

The difference (S -T) there is approx 229.81

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=$\int_{2}^{100} x \log_{2}x dx$

initially.

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

$\int_2^4 x\log_2 x\ dx= 11$

we compute the actual values as:**S**

**T**

where it is seen that **S > T**