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Integrals

Definite Integral

From Approximation to Exact Area

In the Riemann Sums material, we learned how to approximate the area under a curve by dividing it into many rectangles. We also know that the more rectangles we use (the larger the value of nnn), the more accurate our area approximation becomes.

Now, imagine if we could divide the area into an infinite number of rectangles. The width of each rectangle (Δx\Delta xΔx) would become infinitesimally small, approaching zero. This process of taking the limit as the number of partitions nnn approaches infinity is what transforms the Riemann Sum from a mere approximation into an exact calculation. This very concept gives rise to the Definite Integral.

Notation and Meaning of the Definite Integral

The definite integral is the formal way of writing this "infinite sum" of the areas of infinitesimally small rectangles. Mathematically, the definite integral is defined as the limit of a Riemann Sum.

∫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∫ab​f(x)dx=n→∞lim​i=1∑n​f(xi​)Δx

This notation has a specific meaning:

  • ∫ab\int_{a}^{b}∫ab​: This is the integral symbol with a lower limit aaa and an upper limit bbb. These numbers define the interval over which we are calculating the area.
  • f(x)f(x)f(x): This is the integrand, which is the function whose curve we are finding the area under.
  • dxdxdx: Just as with indefinite integrals, this indicates that we are integrating with respect to the variable xxx.

Unlike an indefinite integral, which results in a function (F(x)+CF(x) + CF(x)+C), the result of a definite integral is a single number that represents the net area under the curve from x=ax=ax=a to x=bx=bx=b.

Calculating the Definite Integral with Limits

Let's try to calculate the exact value of a definite integral using its limit definition, as in the reference image example.

Problem: Determine the value of ∫07x dx\int_{0}^{7} x \, dx∫07​xdx.

Solution:

To solve this, we will convert it back to the limit form of a Riemann Sum.

Step 1: Determine the Riemann Sum components

From the problem ∫07x dx\int_{0}^{7} x \, dx∫07​xdx, we know:

  • Function: f(x)=xf(x) = xf(x)=x
  • Interval: [a,b]=[0,7][a, b] = [0, 7][a,b]=[0,7]

Thus, the width of each subinterval is:

Δx=b−an=7−0n=7n\Delta x = \frac{b-a}{n} = \frac{7-0}{n} = \frac{7}{n}Δx=nb−a​=n7−0​=n7​

We will use the right endpoint (xix_ixi​) as the sample point for each partition:

xi=a+iΔx=0+i(7n)=7inx_i = a + i \Delta x = 0 + i \left(\frac{7}{n}\right) = \frac{7i}{n}xi​=a+iΔx=0+i(n7​)=n7i​

Step 2: Set up the Riemann Sum

The height of each rectangle is f(xi)f(x_i)f(xi​), so:

f(xi)=xi=7inf(x_i) = x_i = \frac{7i}{n}f(xi​)=xi​=n7i​

Now, we plug this into the Riemann Sum formula:

∑i=1nf(xi)Δx=∑i=1n(7in)(7n)=∑i=1n49in2\sum_{i=1}^{n} f(x_i) \Delta x = \sum_{i=1}^{n} \left(\frac{7i}{n}\right) \left(\frac{7}{n}\right) = \sum_{i=1}^{n} \frac{49i}{n^2}i=1∑n​f(xi​)Δx=i=1∑n​(n7i​)(n7​)=i=1∑n​n249i​

Step 3: Simplify and use summation properties

We can factor constants out of the sigma notation, as nnn is treated as a constant in the sum from i=1i=1i=1 to nnn.

49n2∑i=1ni\frac{49}{n^2} \sum_{i=1}^{n} in249​i=1∑n​i

Next, we replace ∑i=1ni\sum_{i=1}^{n} i∑i=1n​i with its series summation formula, n(n+1)2\frac{n(n+1)}{2}2n(n+1)​, and then simplify the expression to make it easier to evaluate the limit.

49n2⋅n(n+1)2\frac{49}{n^2} \cdot \frac{n(n+1)}{2}n249​⋅2n(n+1)​
=49n(n+1)2n2= \frac{49n(n+1)}{2n^2}=2n249n(n+1)​
=49(n+1)2n= \frac{49(n+1)}{2n}=2n49(n+1)​
=492(n+1n)=492(1+1n)= \frac{49}{2} \left( \frac{n+1}{n} \right) = \frac{49}{2} \left( 1 + \frac{1}{n} \right)=249​(nn+1​)=249​(1+n1​)

The simplification steps above ensure we get the easiest form to evaluate the limit. First, we cancel nnn from the numerator and denominator, then we split the fraction.

To solve the limit of a Riemann Sum, we often need some common series summation formulas:

  • ∑i=1ni=n(n+1)2\sum_{i=1}^{n} i = \frac{n(n+1)}{2}∑i=1n​i=2n(n+1)​
  • ∑i=1ni2=n(n+1)(2n+1)6\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}∑i=1n​i2=6n(n+1)(2n+1)​
  • ∑i=1ni3=(n(n+1)2)2\sum_{i=1}^{n} i^3 = \left( \frac{n(n+1)}{2} \right)^2∑i=1n​i3=(2n(n+1)​)2

Step 4: Take the Limit

Finally, we take the limit as n→∞n \to \inftyn→∞ to find the exact area.

∫07x dx=lim⁡n→∞492(1+1n)\int_{0}^{7} x \, dx = \lim_{n \to \infty} \frac{49}{2} \left( 1 + \frac{1}{n} \right)∫07​xdx=n→∞lim​249​(1+n1​)
=492(1+0)= \frac{49}{2} \left( 1 + 0 \right)=249​(1+0)
=492= \frac{49}{2}=249​

So, the exact area of the region under the curve f(x)=xf(x) = xf(x)=x from x=0x=0x=0 to x=7x=7x=7 is 49/2 or 24.5. This is an example of how the definite integral provides an exact answer, no longer an approximation.

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Riemann Sum

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Properties of Definite Integral

  • Definite IntegralTransform Riemann sums into exact area calculations using definite integrals. Learn limit processes and notation with detailed worked examples.
On this page
  • From Approximation to Exact Area
  • Notation and Meaning of the Definite Integral
  • Calculating the Definite Integral with Limits
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