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Setting up the limit definition of the definite integral

Consider the function f(x)=xf(x) = \sqrt{x} on the interval [0,4][0, 4]. We want to express the exact area under the curve as a limit of right-endpoint Riemann sums, without evaluating the limit.

Divide [0,4][0,4] into nn subintervals of equal width to form a right-endpoint Riemann sum for f(x)=xf(x) = \sqrt{x}.

python-graph placeholder\begin{aligned} \text{python-graph placeholder} \end{aligned} Setting up the limit definition of the definite integral graph

Let Δx=4−0n\Delta x = \dfrac{4-0}{n} and let xi=0+iΔxx_i = 0 + i\Delta x be the right endpoint of the ii-th subinterval, for i=1,2,…,ni = 1, 2, \dots, n. Write the definite integral ∫04x dx\displaystyle\int_0^4 \sqrt{x}\, dx as a limit of a Riemann sum, expressing Δx\Delta x and f(xi)f(x_i) explicitly in terms of nn and ii. You do not need to evaluate the limit.

Hint

Start with the general formula ∫abf(x) dx=lim⁡n→∞∑i=1nf(xi)Δx\int_a^b f(x)\,dx = \lim_{n\to\infty}\sum_{i=1}^n f(x_i)\Delta x, then substitute a=0a=0, b=4b=4, Δx=4/n\Delta x = 4/n, and xi=4i/nx_i = 4i/n into f(x)=xf(x) = \sqrt{x}.

Step-by-step solution

Since [0,4][0,4] is divided into nn equal subintervals, the width of each subinterval is Δx=4−0n=4n\Delta x = \dfrac{4-0}{n} = \dfrac{4}{n}. The right endpoint of the ii-th subinterval is xi=0+iΔx=4inx_i = 0 + i\Delta x = \dfrac{4i}{n}.

Substituting into f(xi)=xif(x_i) = \sqrt{x_i} gives f(xi)=4inf(x_i) = \sqrt{\dfrac{4i}{n}}. Plugging into the general Riemann sum definition yields ∫04x dx=lim⁡n→∞∑i=1n4in⋅4n.\int_0^4 \sqrt{x}\, dx = \lim_{n\to\infty} \sum_{i=1}^n \sqrt{\dfrac{4i}{n}} \cdot \dfrac{4}{n}. This expression is the required limit form; no further evaluation is needed.

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