Lesson Explainer: Integration by Substitution: Definite Integrals | Nagwa Lesson Explainer: Integration by Substitution: Definite Integrals | Nagwa

Lesson Explainer: Integration by Substitution: Definite Integrals Mathematics

In this explainer, we will learn how to use integration by substitution for definite integrals.

Let us first recall how to integrate by substitution by going through a quick example. Let 𝐼=𝑥2𝑥+5𝑥.d

We compute this integral by making the substitution 𝑢=2𝑥+5. We have that dd𝑢𝑥=6𝑥, and so 𝐼=162𝑥+56𝑥𝑥=16𝑢𝑢𝑥𝑥=16𝑢𝑢.ddddd

In this form, the integral is straightforward to compute: 16𝑢𝑢=16×15𝑢+=130𝑢+.dCC

Finally, we substitute 𝑢=2𝑥+5 back in: 𝐼=130𝑢+=1302𝑥+5+.CC

In general, if we have a function of a function 𝑓(𝑔(𝑥)) and an integral of the form 𝑓(𝑔(𝑥))𝑔𝑥𝑥ddd, then the substitution 𝑢=𝑔(𝑥) yields dddd𝑢𝑥=𝑔𝑥, and so 𝑓(𝑔(𝑥))𝑔𝑥𝑥=𝑓(𝑢)𝑢.dddd

This technique works for definite integrals just as it does for indefinite ones; however, when we make a substitution, the limits of integration need to change.

Formula: Integration by Substitution for Definite Integrals

If 𝑓 is a continuous function and 𝑔 is differentiable with continuous derivative, then 𝑓(𝑔(𝑥))𝑔𝑥𝑥=𝑓(𝑢)𝑢,()()dddd where 𝑢=𝑔(𝑥).

Let us look at an example of using substitution to evaluate a definite integral.

Example 1: Finding the Value of a Definite Integral Using Substitution

Use the substitution 𝑢=𝑥4 to evaluate the definite integral 𝑥(𝑥4)𝑥d.

Answer

Recall that if we are integrating a function of the form 𝑓(𝑔(𝑥))𝑔𝑥dd, then we can make the substitution 𝑢=𝑔(𝑥): 𝑓(𝑔(𝑥))𝑔𝑥𝑥=𝑓(𝑢)𝑢.()()dddd

Observe that after substituting the limits of integration change from 𝑎 and 𝑏 to 𝑔(𝑎) and 𝑔(𝑏).

We want to evaluate 𝑥(𝑥4)𝑥d. Taking 𝑢=𝑥4, we have the following:

  • dd𝑢𝑥=1
  • 𝑥=𝑢+4
  • the limits of integration 𝑢(1)=14=3 and 𝑢(3)=34=1

Thus, 𝑥(𝑥4)𝑥=(𝑢+4)𝑢𝑢=𝑢+4𝑢𝑢=14𝑢+43𝑢=14(1)+43(1)14(3)+43(3)=443.ddd

If we are not told which substitution to use, we will have to choose a suitable substitution ourselves.

Example 2: Finding the Value of a Definite Integral Using Substitution

Find 𝑥𝑥+5𝑥d to the nearest thousandth.

Answer

Recall that if we have a function of the form 𝑓(𝑔(𝑥))𝑔𝑥dd, then we can make the substitution 𝑢=𝑔(𝑥): 𝑓(𝑔(𝑥))𝑔𝑥𝑥=𝑓(𝑢)𝑢.()()dddd

Observe that after substituting the limits of integration change from 𝑎 and 𝑏 to 𝑔(𝑎) and 𝑔(𝑏).

In making a substitution, we are looking for two things. The first thing is an expression, call it 𝑔(𝑥), that appears in the integrand and whose derivative, dd𝑔𝑥, also appears in the integrand. It is worth remembering that if 𝑔(𝑥) is linear in 𝑥 (i.e., 𝑔(𝑥) is of the form 𝑎𝑥+𝑏), then its derivative is a constant; dd𝑔𝑥=𝑎. This means that a linear substitution can always be made.

The second thing we are looking for is a substitution that will make the integral easier to evaluate. When the integrand contains a function of a function, the inner function is often a good candidate for substitution.

These two considerations suggest that 𝑢=𝑥+5 would be a good substitution to try. Indeed, taking 𝑢=𝑥+5, we have the following:

  • dd𝑢𝑥=1
  • 𝑥=𝑢5
  • the limits of integration 𝑢(1)=1+5=4 and 𝑢(4)=4+5=9

Thus, 𝑥𝑥+5𝑥=(𝑢5)𝑢𝑢=(𝑢5)𝑢𝑢=𝑢5𝑢𝑢=25𝑢103𝑢=25910392541034=25×243103×2725×32+103×8=42251903=31615.dddd

Therefore, the value of our definite integral to the nearest thousandth is 21.067.

The method of substitution is often an effective way of tackling trigonometric integrals, as we will see in the next example.

Example 3: Finding the Value of a Definite Integral Using Substitution

Find 9(𝑧)(𝑧)𝑧tansecd.

Answer

Recall that if we have a function of the form 𝑓(𝑔(𝑥))𝑔𝑥dd, then we can make the substitution 𝑢=𝑔(𝑥): 𝑓(𝑔(𝑥))𝑔𝑥𝑥=𝑓(𝑢)𝑢.()()dddd

Observe that after substituting the limits of integration change from 𝑎 and 𝑏 to 𝑔(𝑎) and 𝑔(𝑏).

In making a substitution, we are looking for an expression, call it 𝑔(𝑥), that appears in the integrand and whose derivative, dd𝑔𝑥, also appears in the integrand. Now, trigonometric functions differentiate to other trigonometric functions. For this reason, when presented with an integral involving multiple trigonometric functions, it is worth checking to see if any of them are the derivatives of other ones.

We want to evaluate 9(𝑧)(𝑧)𝑧tansecd. Taking 𝑢=(𝑧)tan, we have the following:

  • ddsec𝑢𝑧=(𝑧)
  • the limits of integration 𝑢(0)=(0)=0tan and 𝑢𝜋4=𝜋4=1tan

Thus, 9(𝑧)(𝑧)𝑧=9𝑢𝑢=92𝑢=921920=92.tansecdd

Recall that the derivative of ln(𝑥) is 1𝑥. This means that quotients with ln(𝑥) in the numerator are often good candidates for integration by substitution, as in the next example.

Example 4: Finding the Value of a Definite Integral Using Substitution

Determine 𝑥𝑥𝑥lnd.

Answer

Recall that if we have a function of the form 𝑓(𝑔(𝑥))𝑔𝑥dd, then we can make the substitution 𝑢=𝑔(𝑥): 𝑓(𝑔(𝑥))𝑔𝑥𝑥=𝑓(𝑢)𝑢.()()dddd

Observe that after substituting the limits of integration change from 𝑎 and 𝑏 to 𝑔(𝑎) and 𝑔(𝑏).

Since our integrand in this case consists of a fraction with ln(𝑥) in the numerator, 𝑢=(𝑥)ln is an obvious candidate for substitution. Indeed, taking 𝑢=(𝑥)ln, we have the following:

  • dd𝑢𝑥=1𝑥
  • the limits of integration 𝑢(1)=(1)=0ln and 𝑢(𝑒)=1

Thus, 𝑥𝑥𝑥=𝑢𝑢=12𝑢=12×112×0=12.lndd

Let us look at a more complicated example of using substitution to evaluate a definite integral involving the natural logarithm.

Example 5: Finding the Value of a Definite Integral Using Substitution

Evaluate 5𝑥7(𝑥7)+1𝑥lnd.

Answer

Recall that if we have a function of the form 𝑓(𝑔(𝑥))𝑔𝑥dd, then we can make the substitution 𝑢=𝑔(𝑥): 𝑓(𝑔(𝑥))𝑔𝑥𝑥=𝑓(𝑢)𝑢.()()dddd

Observe that after substituting the limits of integration change from 𝑎 and 𝑏 to 𝑔(𝑎) and 𝑔(𝑏).

Although the integrand in this example looks rather complicated, we should immediately notice that it contains a natural logarithm and a fraction whose denominator is closely related to the argument of that logarithm. By using the chain rule, we get ddln𝑥(𝑥7)=2(𝑥7)(𝑥7)=2𝑥7.

This suggests that 𝑢=(𝑥7)ln might be an extremely good candidate for substitution. However, observe that the constant term in ln(𝑥7)+1 differentiates to 0, so we may as well take 𝑢=(𝑥7)+1ln. With 𝑢=(𝑥7)+1ln, we have the following:

  • dd𝑢𝑥=2𝑥7
  • the limits of integration 𝑢(8)=(1)+1=1ln and 𝑢(𝑒+7)=𝑒+1=2+1=3ln

Thus, 5𝑥7(𝑥7)+1𝑥=52𝑢𝑢,=58𝑢=58×358×1=5×(811)8=4008=50.lndd

Finally, let us look at an example combining an exponential function and trigonometric functions.

Example 6: Finding the Value of a Definite Integral Using Substitution

Evaluate 𝑒𝜃𝜃sincosd.

Answer

Recall that if we have a function of the form 𝑓(𝑔(𝑥))𝑔𝑥dd, then we can make the substitution 𝑢=𝑔(𝑥): 𝑓(𝑔(𝑥))𝑔𝑥𝑥=𝑓(𝑢)𝑢.()()dddd

Observe that after substituting the limits of integration change from 𝑎 and 𝑏 to 𝑔(𝑎) and 𝑔(𝑏). Consider the expression 𝑒𝜃𝜃.sincosd

Notice the presence of a sine function in the exponent and right next to it its derivative, a cosine. Indeed, with 𝑢=(𝜃)sin, we have ddcos𝑢𝜃=𝜃, and so 𝑒𝜃𝜃=𝑒𝑢.()()sincosdd

We have 𝑢𝜋3=𝜋3=32sin and 𝑢(𝜋)=(𝜋)=0sin. Therefore, 𝑒𝜃𝜃=𝑒𝑢=12𝑒=1212𝑒=121𝑒.()()sincosdd

Let us finish by recapping a few important concepts from this explainer.

Key Points

  • We can identify situations where a substitution can be used to simplify a definite integral.
  • In particular, we can recognize expressions of the form 𝑓(𝑔(𝑥))𝑔𝑥,dd which are the product of a function of a function 𝑓(𝑔(𝑥)) and the derivative dd𝑔𝑥 of the inner function.
  • In this situation, we can make the substitution 𝑢=𝑔(𝑥); that is, 𝑓(𝑔(𝑥))𝑔𝑥𝑥=𝑓(𝑢)𝑢.()()dddd
  • We can evaluate such definite integrals, taking care that when we make a substitution 𝑢=𝑔(𝑥), the limits of integration change: 𝑎𝑔(𝑎)𝑏𝑔(𝑏).

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