In everyday life, we often need to calculate the area of various flat shapes. For shapes with simple forms like squares or triangles, we can use familiar formulas. But what if we want to calculate the area of a region bounded by irregular curves?
Definite integrals provide an elegant solution to this problem. The basic concept of definite integrals stems from Riemann's approach, where we divide a region into small rectangles and then sum their areas.
Imagine we have a function f(x) and want to find the area of the region under the curve from x=a to . We can divide the interval into small parts with width .
The first step is to determine the lower and upper limits of integration. These limits indicate the range of x values that bound the region whose area we want to calculate.
Let's apply this concept to a concrete example. Suppose we want to calculate the area of the region bounded by the curve f(x)=x2−4x and the x-axis between x=1 and x=3.
Graph of Function f(x)=x2−4x
Visualization of the region whose area will be calculated with the help of boundary lines and shaded area.
Now, try to observe the graph above. The function f(x)=x2−4x turns out to have negative values in the interval [1,3]. We can easily check: when x=1, we get f(1)=1−4=−3. Similarly, when x=3, we get f(3)=9−12=−3.
Well, here's where the uniqueness lies! Since we're looking for area which is always positive, we need to use the absolute value of that function. So our integral becomes:
A=∫13∣x2−4x∣dx
A=∫13−(x2−4x)dx
A=∫13(−x2+4x)dx
Let's solve it step by step:
A=[−3x3+2x2]13
A=(−327+18)−(−31+2)
A=9−35=322
Therefore, the area of that region is 322 square units.
Now let's try a slightly more challenging example with an irrational function. We will calculate the area of the region under the curve f(x)=xx2+5 from x=0 to x=2.
Graph of Function f(x)=xx2+5
Region under the curve whose area will be calculated with interval helper lines.
For this integral, we need to use the substitution technique. Why? Because there's the form xx2+5 which is quite complex if we solve it directly.
Let's perform substitution with u=x2+5. From here, we get the differential du=2xdx, which means xdx=21du.
Don't forget to change the integration limits too! When x=0, we get u=5. When x=2, we get u=9.
Now our integral becomes:
A=∫02xx2+5dx
A=∫5921udu
A=21∫59u1/2du
A=21[32u3/2]59
A=31[u3/2]59
A=31(27−55)
Note that 93/2=(32)3/2=33=27 and 53/2=5⋅5.
When using substitution in definite integrals, don't forget to change the integration limits according to the new substitution variable.
