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The familiar formulas for rectangles and triangles do not apply directly to a region with a curved boundary.
Definite integrals handle such regions by accumulating the areas of increasingly narrow strips. A Riemann sum partitions the interval, approximates each strip with a rectangle, and takes the limit of their total area.
Suppose a function bounds a region from to . We divide into subintervals of width .
Use these three steps to calculate the area:
The first step is to determine the lower and upper limits of integration. These limits indicate the range of values that bound the region whose area we want to calculate.
The integrand describes the vertical height of the region. If throughout , the area is . If the graph crosses the axis, split the interval at its zeros or integrate so that every part contributes positive area.
After determining the limits and function, we can evaluate the integral using the fundamental theorem of calculus:
In this formula, is an antiderivative of .
Consider the region bounded by , the -axis, , and .
On , factor the function as . Every point in the interval satisfies and , so throughout the interval. The endpoint values and agree with this sign analysis.
Geometric area is nonnegative. Because the function is negative throughout this interval, its absolute value equals its negative, so:
Now evaluate the definite integral:
The area of that region is square units.
For a second example, calculate the area under from to .
The expression suggests substitution because the factor is proportional to the derivative of the expression inside the square root.
Set . Then , so .
The limits must use the same variable as the integrand: gives , while gives .
Now our integral becomes:
Here, and .
When using substitution in definite integrals, don't forget to change the integration limits according to the new substitution variable.
Calculate the area of the region bounded by the curve , the -axis, and the lines and !
Determine the area of the region under the curve from to !
Calculate the area of the region bounded by the curve and the -axis!
First problem with function
Since this function is always positive, we can directly set up the integral:
After we integrate and evaluate, we obtain:
Therefore, the area of that region is square units.
Second problem with rational function
The antiderivative used in this integral is because its derivative is .
Third problem with parabola
First, find where the curve intersects the -axis:
Published: . Updated: .
The area of that region is square units.
So the intersection points are at and . Since this function is positive between these two points, we can directly integrate:
The area of that region is square units.
