In this section we learn to compute the value of a definite integral using the fundamental theorem of calculus.
1 The Fundamental Theorem of Calculus
We begin by recalling the definition of the definite integral: where and is a sample point for the -th sub-interval, .
Computing the value of a definite integral from this definition can be cumbersome and require knowledge of special summation formulas. Fortunately, there is a theorem which states that computing definite integrals can done via anti-differentiation!
The proof of the Fundamental Theorem of Calculus can be obtained by applying the Mean Value Theorem to on each of the sub-intervals and using the value of in each case as the sample point.
The process of calculating the numerical value of a definite integral is performed in two main steps: first, find the anti-derivative and second, plug the endpoints of integration, and to compute . To symbolize the steps, we use a vertical evaluation bar:
In the first step, the function is introduced and in the second step, the difference of the values and is computed.
In the following examples, we compute definite integrals using the FTC in order to solve the area problem. Recall that the definite integral of a non-negative function gives the area under the curve.
Since the function is non-negative on the interval , the exact area under the curve is the definite integral We can compute the value of this definite integral using the Fundamental Theorem of Calculus as follows:
Since the function is non-negative on the interval , the exact area can be expressed as the definite integral We can use the Fundamental Theorem of Calculus to compute the value of this definite integral as follows:
Since the function is non-negative on the interval , the area can be expressed as the definite integral We can use the Fundamental Theorem of Calculus to compute the value of this definite integral as follows:
Since the function is non-negative on the interval , the exact area can be expressed as the definite integral We can use the Fundamental Theorem of Calculus to compute the value of this definite integral as follows:
Since the function is non-negative on the interval , the exact area can be expressed as the definite integral We can use the Fundamental Theorem of Calculus to compute the value of this definite integral as follows:
Since the function is non-negative on the interval , the exact area can be expressed as the definite integral We can use the Fundamental Theorem of Calculus to compute the value of this definite integral as follows:
Since the function is non-negative on the interval , the exact area can be expressed as the definite integral We can use the Fundamental Theorem of Calculus to compute the value of this definite integral as follows:
We now consider the definite integral of negative functions. If for all in the interval , then for any Riemann Sum of on . Hence, the definite integral, as well. How should we interpret the definite integral in this case? The graph of will lie below the -axis, and the definite integral will equal times the area of the region above the curve.
To put it another way, if on then
Since the function is negative (or zero) on the interval , the exact area above the curve is the negative of the definite integral We can compute the value of this definite integral using the Fundamental Theorem of Calculus as follows:
Hence the area of the region above the curve is .
Since the function on the interval , the exact area above the curve is the negative of the definite integral We can compute the value of this definite integral using the Fundamental Theorem of Calculus as follows:
Hence the area of the region above the curve is .
Since the function on the interval , the exact area above the curve is the negative of the definite integral We can compute the value of this definite integral using the Fundamental Theorem of Calculus as follows:
Hence the area of the region above the curve is .
2 More Definite Integrals
3 Definite Integrals and Substitution
We now compute definite integrals that require a u-substitution. The key is to change the limits of integration when we change the variable. Suppose that Now, consider the definite integral We will make the substitution, , so that . When we convert to the integral in terms of the variable , we will change the limits of integration from and to and respectively. This will give us In this way, we do not need to go basck to the variable , as we did with a substitution in an indefinite integral.