Carmen has two master's degrees in mathematics has has taught mathematics classes at the college level for the past 9 years.
In this lesson, we will discuss changing the limits of definite integrals. This is useful when using u-substitution or trigonometric substitution to solve definite integrals and saves time as there is no need to revert back to the original variable.
Expressing the Same Idea
You are new to town and ask a couple friends for directions to the park. One tells you to drive north for 2 miles, then drive west for 1 mile. The other tells you to drive north for about 4 minutes, then turn left and drive for about 2 minutes. Both friends get you to the park. Is one way better than the other? No! Some people prefer to measure precise distances, and others are better off with time. In a similar manner, the idea of changing the limits of definite integrals when substituting is just expressing the same problem in a different way.
When evaluating an integral using u-substitution, an expression involving the original variable is replaced by a new variable. The end result is a simpler integral.
When we use u-substitution to evaluate a definite integral, we must change each part of the integral to use the new variable (usually u) instead of the old variable (often x). There are three pieces that must be changed:
The function itself (change an expression involving x into u)
The differential (change dx into an expression involving du)
The bounds (change the bounds of integration from values for x into values for u)
Hopefully, the first two are familiar to you. We will focus on the third step. To change the bounds, use the expression that relates x and u. Plug in the original lower bound for x and solve for u. This gives the new lower bound. Then plug in the original upper bound for x and solve for u to find the new upper bound.
Evaluate the definite integral ∫ √(4x+1)dx from 0 to 2.
What do you do first? As with most definite integrals, you should ignore the bounds (0 and 2) at first and focus on how to find an antiderivative of the function inside the integral. We need an antiderivative of √(4x+1).
This one is simple enough that you might be able to do it without using u-substitution, but especially for beginning students, it might be easier to substitute u = 4x + 1. Then, take the derivative to see that du = 4dx, and solve for dx to get dx = 1/4 * du.
Here is the new part. We have changed all the expressions involving x to expressions involving u. Remember that the bounds 0 and 2 on the integral were bounds for x, not u! So now we can change those also.
Remember that u = 4x + 1. That means that when x = 0, u = 4(0) + 1 = 1, and when x = 2, u = 4(2) + 1 = 9. So, while the old bounds for x were 0 and 2, the new bounds for u are 1 and 9. Putting all these pieces together, the original integral is rewritten as follows:
∫ √(4x+1)dx from 0 to 2 = ∫ √u * 1/4 * du from 1 to 9.
Do you see how everything that talked about x (the x itself, the dx, and the bounds) has been changed into talking about u instead? That means that we can now forget about x completely and just evaluate the integral using the variable u.
Finishing up, an antiderivative for √u is 2/3*u3/2, so we can finish:
∫ √u * 1/4 * du from 1 to 9
= 2/3 * u3/2 * 1/4 evaluated from 1 to 9
= 1/6(93/2 - 13/2)
= 1/6(27 - 1)
Evaluate the definite integral ∫ sin3x cosxdx from π/3 to π/2.
You should notice that the function sin3x cosx is pretty complicated, so it will be helpful to try a u-substitution to simplify the expression.
Let's use u = sinx. Then du = cosxdx, so the integral now has u3du in it, and all that remains is to change the bounds.
Since u = sinx, when x = π/3, u = sin(π/3) = √(3)/2. Also, when x = π/2, u = sin(π/2) = 1. Our integral is rewritten as follows:
∫ sin3x cosxdx from π/3 to π/2
= ∫ u3du from √(3)/2 to 1
We now solve it as usual, using the variable u and forgetting that the variable x ever existed.
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The other type of substitution often used to solve integrals is trigonometric substitution. In this type of substitution, we substitute for x a trigonometric function of θ. This substitution simplifies the integral into something we can handle. This technique is one of the most challenging ones in integral calculus, so if you are a bit confused, you are not alone!
Evaluate the definite integral ∫ (dx / √(x2 + 9) ) from 0 to √3.
Notice, that we cannot use any 'simple' methods to find an antiderivative for this function, so we have no choice but to use a more complicated method. Based on the fact that we see x2 + 9 in this integral, we will use the substitution x = 3tanθ.
To find dx, we take the derivative of x = 3tanθ to get dx = 3sec2θdθ. We also need to simplify the square root expression inside the integral:
√(x2 + 9)
= √( (3tanθ)2 + 9)
= √(9tan2θ + 9)
= √( 9(tan2θ + 1) )
With this simplification and the expression we find for dx, the expression inside the integral simplifies to:
dx / √(x2 + 9)
= 3sec2θdθ / (3secθ)
Much simpler! That's the power of trigonometric substitution!
Now we only have to worry about changing the limits. We do this by using the initial relationship x = 3tanθ and the initial limits for x: 0 and √3. First, we plug in 0 for x and solve for θ using the arctan function:
0 = 3tanθ
0 = tanθ
θ = arctan(0)
θ = 0
Now plug in √3 and solve for θ:
√3 = 3tanθ
√(3)/3 = tanθ
θ = arctan( √(3)/3 )
θ = π/6
Putting all the pieces together, we rewrite the original integral and solve using the new variable θ. Notice that we have changed the function itself, the dx, and the limits of integration to expressions for θ rather than x, so we can forget about the x when we actually integrate.
∫ (dx / √(x2 + 9) ) from 0 to √3
= ∫ secθdθ from 0 to π/6
= ln |secθ + tanθ| evaluated from 0 to π/6
= ln |sec(π/6) + tan(π/6)| - ln |sec(0) + tan(0)|
= ln |2/√3 + 1/√3| - ln |1 + 0|
= ln(3/√3) - ln(1)
= ln(√3) - 0
When solving a definite integral using either u-substitution or trigonometric substitution, we may change the bounds of integration. To do this, remember to switch all three aspects of the integral to refer to the new variable: the function itself, the differential (dx), and the limits. Once all three pieces have been changed, you can evaluate the integral using the new variable and completely forget about the old variable.
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