Q. 12 - Calculus

Question

Riemann sums: Calculate each of the following Riemann sum

approximations for the definite integral of f on [a, b], using the

given value of n.

The lower sum for $f\left(x\right)=4x-x^2$ on $\left[0,3\right]$, $n=6$ .

Step-by-Step Solution

Answer

The lower sum is 5 .

Step 1. Given information .

Consider the given function $f\left(x\right)=4x-x^2$ on $\left[0,3\right]$ .

Step 2. Formula used .

Riemann sum formula for lower sum is $L\left(P,f\right)=\sum _{r=0}^3{m}_r\delta x_r$ .

Step 3. Find the lower sum for f x = 4 x - x 2 on 0 , 3 .

Divide the no in open interval from zero to 3 in 6 sub interval to find ${\delta }_x$

and m . The sub intervals are $\left(0,1\right) ,\left(1,2\right) ,\left(2,3\right) ,\left(3,4\right) ,\left(4,5\right) ,\left(5,6\right)$ .

The value of $m_1=0 , m_2=1 ,m_3=2 , m_4=3 , m_5=4 , m_6=5$

Put the values of m in given function $f\left(x\right)=4x-x^2$. we get $f\left(x\right)=4x-x^2=4\times 0-0=0=m_1$

$m_2=3 , m_3=4 , m_4=3 , m_5=0 , m_6=-5$

Substitute the values in formula $L\left(P,f\right)=\sum _{r=0}^3{m}_r\delta x_r$

where $\delta x$ is 1.

$$\begin{gathered} \Rightarrow m_1\delta x_1+m_2\delta x_2+m_3\delta x_3+m_4\delta x_4+m_5\delta x_5+m_6\delta x_6 \\ \Rightarrow 0\times 1+3\times 1+4\times 1+3\times 1+0\times 1+\left(-5\right)\times 1 \\ \Rightarrow 0+3+4+3+0-5 \\ \Rightarrow 5 \end{gathered}$$