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Setting up the Riemann sum limit for the area under f

Consider the piecewise function f(x)={xif x≤26−xif x>2f(x) = \begin{cases} x & \text{if } x \le 2 \\ 6-x & \text{if } x > 2 \end{cases} on the interval [0,4][0,4].

The graph of f(x)f(x) on [0,4][0,4] forms a triangular region above the xx-axis (rising from (0,0)(0,0) to (2,2)(2,2), then falling to (4,2)(4,2)). Using a partition of [0,4][0,4] into nn subintervals of equal width, with xi∗x_i^* denoting a sample point in the ii-th subinterval, write the exact expression (as a limit of a Riemann sum) that defines ∫04f(x) dx.\int_0^4 f(x)\,dx. Be sure to state Δx\Delta x in terms of nn explicitly.

Setting up the Riemann sum limit for the area under f graph
Hint

Recall the general form ∫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. Identify aa, bb, and Δx=(b−a)/n\Delta x = (b-a)/n; you do not need to split the sum into two pieces or evaluate it — just write down the correct limit expression using ff as given.

Step-by-step solution

Since the interval is [0,4][0,4], the width of each subinterval is Δx=4−0n=4n\Delta x = \frac{4-0}{n} = \frac{4}{n}. The definition of the definite integral as a limit of Riemann sums applies directly here, regardless of the fact that ff is piecewise-defined, since ff is still a single function on [0,4][0,4].

Substituting a=0a=0, b=4b=4, and Δx=4/n\Delta x = 4/n into the general definition gives ∫04f(x) dx=lim⁡n→∞∑i=1nf(xi∗)⋅4n,\int_0^4 f(x)\,dx = \lim_{n\to\infty}\sum_{i=1}^n f(x_i^*)\cdot \frac{4}{n}, where xi∗x_i^* is any sample point chosen in the ii-th subinterval, and f(xi∗)f(x_i^*) equals xi∗x_i^* if xi∗≤2x_i^*\le 2 or 6−xi∗6-x_i^* if xi∗>2x_i^*>2. This expression represents the area of the triangular region shown, without needing to evaluate the limit.

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