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 $\left[0,3\right]$,$n=6$.

Short answer

The lower sum is 5 .

Step-by-step solution

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 fx=4x-x2 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.

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