Answer to Question #287307 in Calculus for Hussain Darboe

Question

1. Evaluate the limit by first recognizing the sum as a Reimann Sum for a function defined on [0,1] limit as n tends to infinity 1/n (the root of 1/n + root of 2/n + root of 3/n +....+ root of n/n)

2. If f prime is continuous on [a,b], show that 2* the integral from a to b of f(x) f prime(x) dx= f(b) square minus f(a) square.

Answer 1

Solution:

1.

$\begin{aligned} &\lim _{n \rightarrow \infty} \frac{1}{n}\left(\sqrt{\frac{1}{n}}+\sqrt{\frac{2}{n}}+\sqrt{\frac{3}{n}}+\ldots+\sqrt{\frac{n}{n}}\right) \\ &=\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{i=1}^{n} \sqrt{\frac{i}{n}} \\ &=\lim _{n \rightarrow \infty} \sum_{i=1}^{n} \sqrt{\frac{i}{n}} \cdot \frac{1}{n} \\ &=\lim _{n \rightarrow \infty} \sum_{i=1}^{n} \sqrt{0+i \Delta x} \cdot \Delta x \\ &\text { Where } \Delta x=\frac{1}{n} \\ &\text { Note that } \lim _{n \rightarrow \infty} \sum_{i=1}^{n} f(a+i \Delta x) \cdot \Delta x=\int_{a}^{b} f(x) d x \\ &\text { Here } a=0 \\ &\text { and } \Delta x=\frac{b-a}{n}=\frac{1}{n} \\ &\text { Therefore, } a=0 \Rightarrow b=1\\ &f(x)=\sqrt{x} \end{aligned}$

So, $\int_{a}^{b} f(x) d x=\int_{0}^{1} \sqrt x d x$

$=[\dfrac{x^{3/2}}{3/2}]_0^1 \\=\dfrac23(1-0) \\=\dfrac23$

2. We know that,

$\dfrac{d}{dx} f^2(x) = 2f(x)f'(x)$ ...(i)

Now, consider

$2∫_a^b f(x) f'(x) dx \\=∫_a^b\dfrac{d}{dx} f^2(x) \ [Using\ (i)] \\=f^2(x) |_{x=a}^{x=b}$ [Using Fundamental theorem of calculus]

$=f^2(b) - f^2(a).$