Answer to Question #203192 in Real Analysis for Rajkumar

Question

For the function, f(x) = x² −2 defined over [1,5], verify : L(P, f ) ≤ U(−P, f )

where P is the partition which divides [1,5] into four equal intervals.

Answer 1

When P is the position that divides [1,5] into four equal parts. The position of [1,5] into these parts are, $P=[I_1, I_2, I_3, I_4]$

Where I₁=[1,2], I₂=[2,3],I₃=[3,4],I₄=[4,5]

Now the length of each interval is given by

$S_1=2-1=1= \triangle x_1$

$S_2=3-2=1= \triangle x_2$

$S_3=4-3=1= \triangle x_3$

$S_4=5-4=1= \triangle x_4$

Now the downbound of the function f(x) corresponding to each interval is given by

$\therefore f(x)=x^2-2$

$m_1=1^2-2=-1$

$m_2=2^2-2=2$

$m_3=3^2-2=7$

$m_4=4^2-2=14$

$L(p,f)= \sum_{r=1}^4 m_r \triangle x_r =m_1 \triangle x_1+m_2 \triangle x_2+m_3 \triangle x_3+m_4\triangle x_4$

$L(p,f) =-1*1+2*1+7*1+14*1 =-1+2+7+14=22$

Now $-P=[-I_1, -I_2, -I_3, -I_4]$

Where I'₁=[-1,-2], I'₂=[-2,-3],I'₃=[-3,-4],I'₄=[-4,-5]

The length of each interval = $1 = \triangle x_r \space \space \space 1 \eqslantless r \eqslantless 4$

The upperbound of $f(x)=x^2-2$ over each interval is given by

$\therefore f(x)=x^2-2$

$m_1=(-1)^2-2=-1$

$m_2=(-2)^2-2=2$

$m_3=(-3)^2-2=7$

$m_4=(-4)^2-2=14$

The upper sum is defined as

$U(-p,f)= \sum_{r=1}^4 m_r \triangle x_r$

$U(-p,f) =-1*1+2*1+7*1+14*1 =-1+2+7+14=22$

Hence $L(p,f)=U(-p,f)$ verified

We also know that the upper sum is always greater than the lower sum, hence

$L(p,f)\eqslantless U(-p,f)$ verified