Science:Math Exam Resources/Courses/MATH101/April 2005/Question 08 (a)

MATH101 April 2005

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Question 08 (a)

An unknown continuous function $f(x)$ satisfies $f(0)=0$, $f(4)=8$, and $2x\leq f(x)\leq 6x-x^{2}$ for $0\leq x\leq 4.$ Also, $f(x)$ is nondecreasing on this interval, i.e. it satisfies $f(c)\leq f(d)$ for all real numbers $c$ and $d$ with $0\leq c\leq d\leq 4.$ Let $I$ be the value of definite integral $\int _{0}^{4}f(x)\,dx.$ (a) Let $L_{100}$ be the underestimate for $I$ obtained by using a Riemann sum with equal-length subintervals and $x_{i}^{*}=x_{i-1}$ (i.e. using the left endpoints of the subintervals), and $R_{100}$ be the overestimate obtained by using $x_{i}^{*}=x_{i}$ (i.e. the right endpoints). Compute a numerical value for $R_{100}-L_{100}$.

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Hint
The right hand Riemann sum is computed by $R_{n}=\sum _{i=1}^{n}f(x_{i})\Delta x$ whereas the left hand Riemann sum is computed by $L_{n}=\sum _{i=0}^{n-1}f(x_{i})\Delta x$ What happens when you subtract these two values?

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. The right hand Riemann sum is computed by $R_{100}=\sum _{i=1}^{100}f(x_{i})\Delta x$ whereas the left hand Riemann sum is computed by $L_{100}=\sum _{i=0}^{99}f(x_{i})\Delta x$ and hence, we have ${\begin{aligned}R_{100}-L_{100}&=\sum _{i=1}^{100}f(x_{i})\Delta x-\sum _{i=0}^{99}f(x_{i})\Delta x\\&=f(x_{100})\Delta x+\sum _{i=1}^{99}f(x_{i})\Delta x-\sum _{i=1}^{99}f(x_{i})\Delta x-f(x_{0})\Delta x\\&=f(4)\Delta x-f(0)\Delta x\\&=8\Delta x-(0)\Delta x\\&=8\Delta x\end{aligned}}$ By definition, we know $\Delta x={\frac {b-a}{n}}={\frac {4-0}{100}}={\frac {1}{25}}$ and so ${\begin{aligned}R_{100}-L_{100}=8\Delta x={\frac {8}{25}}\end{aligned}}$ completing the proof.

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