Integration by Parts, Part 1
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Transcript of Integration by Parts, Part 1
![Page 1: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/1.jpg)
![Page 2: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/2.jpg)
![Page 3: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/3.jpg)
The Idea of Integration by Parts
![Page 4: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/4.jpg)
The Idea of Integration by Parts
The idea is to use the product rule in reverse.
![Page 5: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/5.jpg)
The Idea of Integration by Parts
The idea is to use the product rule in reverse. Remember theproduct rule?
![Page 6: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/6.jpg)
The Idea of Integration by Parts
The idea is to use the product rule in reverse. Remember theproduct rule?
[u(x)v(x)]′ =?
![Page 7: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/7.jpg)
The Idea of Integration by Parts
The idea is to use the product rule in reverse. Remember theproduct rule?
[u(x)v(x)]′ = u′(x)v(x)
![Page 8: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/8.jpg)
The Idea of Integration by Parts
The idea is to use the product rule in reverse. Remember theproduct rule?
[u(x)v(x)]′ = u′(x)v(x)+
![Page 9: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/9.jpg)
The Idea of Integration by Parts
The idea is to use the product rule in reverse. Remember theproduct rule?
[u(x)v(x)]′ = u′(x)v(x) + u(x)v ′(x)
![Page 10: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/10.jpg)
The Idea of Integration by Parts
The idea is to use the product rule in reverse. Remember theproduct rule?
[u(x)v(x)]′ = u′(x)v(x) + u(x)v ′(x)
Let’s now integrate both sides:
![Page 11: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/11.jpg)
The Idea of Integration by Parts
The idea is to use the product rule in reverse. Remember theproduct rule?
[u(x)v(x)]′ = u′(x)v(x) + u(x)v ′(x)
Let’s now integrate both sides:∫[u(x)v(x)]′ dx =
![Page 12: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/12.jpg)
The Idea of Integration by Parts
The idea is to use the product rule in reverse. Remember theproduct rule?
[u(x)v(x)]′ = u′(x)v(x) + u(x)v ′(x)
Let’s now integrate both sides:∫[u(x)v(x)]′ dx =
∫u′(x)v(x)dx
![Page 13: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/13.jpg)
The Idea of Integration by Parts
The idea is to use the product rule in reverse. Remember theproduct rule?
[u(x)v(x)]′ = u′(x)v(x) + u(x)v ′(x)
Let’s now integrate both sides:∫[u(x)v(x)]′ dx =
∫u′(x)v(x)dx+
![Page 14: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/14.jpg)
The Idea of Integration by Parts
The idea is to use the product rule in reverse. Remember theproduct rule?
[u(x)v(x)]′ = u′(x)v(x) + u(x)v ′(x)
Let’s now integrate both sides:∫[u(x)v(x)]′ dx =
∫u′(x)v(x)dx +
∫u(x)v ′(x)dx
![Page 15: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/15.jpg)
The Idea of Integration by Parts
The idea is to use the product rule in reverse. Remember theproduct rule?
[u(x)v(x)]′ = u′(x)v(x) + u(x)v ′(x)
Let’s now integrate both sides:∫[u(x)v(x)]′ dx =
∫u′(x)v(x)dx +
∫u(x)v ′(x)dx
What happens when you integrate a derivative?
![Page 16: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/16.jpg)
The Idea of Integration by Parts
The idea is to use the product rule in reverse. Remember theproduct rule?
[u(x)v(x)]′ = u′(x)v(x) + u(x)v ′(x)
Let’s now integrate both sides:∫[u(x)v(x)]′ dx =
∫u′(x)v(x)dx +
∫u(x)v ′(x)dx
What happens when you integrate a derivative?∫[u(x)v(x)]′ dx =?
![Page 17: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/17.jpg)
The Idea of Integration by Parts
The idea is to use the product rule in reverse. Remember theproduct rule?
[u(x)v(x)]′ = u′(x)v(x) + u(x)v ′(x)
Let’s now integrate both sides:∫[u(x)v(x)]′ dx =
∫u′(x)v(x)dx +
∫u(x)v ′(x)dx
What happens when you integrate a derivative?∫[u(x)v(x)]′ dx =?
By definition of indefinite integral, you get back the functionyou’re taking the derivative of:
![Page 18: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/18.jpg)
The Idea of Integration by Parts
The idea is to use the product rule in reverse. Remember theproduct rule?
[u(x)v(x)]′ = u′(x)v(x) + u(x)v ′(x)
Let’s now integrate both sides:∫[u(x)v(x)]′ dx =
∫u′(x)v(x)dx +
∫u(x)v ′(x)dx
What happens when you integrate a derivative?∫[u(x)v(x)]′ dx =?
By definition of indefinite integral, you get back the functionyou’re taking the derivative of:∫
[u(x)v(x)]′ dx = u(x)v(x)
![Page 19: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/19.jpg)
The Idea of Integration by Parts
![Page 20: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/20.jpg)
The Idea of Integration by Parts
So, our formula becomes:
![Page 21: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/21.jpg)
The Idea of Integration by Parts
So, our formula becomes:
u(x)v(x) =
∫u′(x)v(x)dx +
∫u(x)v ′(x)dx
![Page 22: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/22.jpg)
The Idea of Integration by Parts
So, our formula becomes:
u(x)v(x) =
∫u′(x)v(x)dx +
∫u(x)v ′(x)dx
Now, if we subtract∫u′(x)v(x)dx to both sides we get:
![Page 23: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/23.jpg)
The Idea of Integration by Parts
So, our formula becomes:
u(x)v(x) =
∫u′(x)v(x)dx +
∫u(x)v ′(x)dx
Now, if we subtract∫u′(x)v(x)dx to both sides we get:
u(x)v(x) −∫
u′(x)v(x)dx =
∫u(x)v ′(x)dx
![Page 24: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/24.jpg)
The Idea of Integration by Parts
So, our formula becomes:
u(x)v(x) =
∫u′(x)v(x)dx +
∫u(x)v ′(x)dx
Now, if we subtract∫u′(x)v(x)dx to both sides we get:
u(x)v(x) −∫
u′(x)v(x)dx =
∫u(x)v ′(x)dx
Or:
![Page 25: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/25.jpg)
The Idea of Integration by Parts
So, our formula becomes:
u(x)v(x) =
∫u′(x)v(x)dx +
∫u(x)v ′(x)dx
Now, if we subtract∫u′(x)v(x)dx to both sides we get:
u(x)v(x) −∫
u′(x)v(x)dx =
∫u(x)v ′(x)dx
Or: ∫u(x)v ′(x)dx = u(x)v(x) −
∫u′(x)v(x)dx
![Page 26: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/26.jpg)
The Idea of Integration by Parts
So, our formula becomes:
u(x)v(x) =
∫u′(x)v(x)dx +
∫u(x)v ′(x)dx
Now, if we subtract∫u′(x)v(x)dx to both sides we get:
u(x)v(x) −∫
u′(x)v(x)dx =
∫u(x)v ′(x)dx
Or: ∫u(x)v ′(x)dx = u(x)v(x) −
∫u′(x)v(x)dx
This is the formula of integration by parts.
![Page 27: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/27.jpg)
Example 1
![Page 28: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/28.jpg)
Example 1
Let’s try to find the integral:
![Page 29: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/29.jpg)
Example 1
Let’s try to find the integral: ∫xexdx
![Page 30: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/30.jpg)
Example 1
Let’s try to find the integral: ∫xexdx
This integral can’t be solved with any of our previous methods.
![Page 31: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/31.jpg)
Example 1
Let’s try to find the integral: ∫xexdx
This integral can’t be solved with any of our previous methods.So, let’s try integration by parts!
![Page 32: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/32.jpg)
Example 1
Let’s try to find the integral: ∫xexdx
This integral can’t be solved with any of our previous methods.So, let’s try integration by parts!Our formula is:
![Page 33: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/33.jpg)
Example 1
Let’s try to find the integral: ∫xexdx
This integral can’t be solved with any of our previous methods.So, let’s try integration by parts!Our formula is:∫
u(x)v ′(x)dx = u(x)v(x) −∫
u′(x)v(x)dx
![Page 34: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/34.jpg)
Example 1
Let’s try to find the integral: ∫xexdx
This integral can’t be solved with any of our previous methods.So, let’s try integration by parts!Our formula is:∫
u(x)v ′(x)dx = u(x)v(x) −∫
u′(x)v(x)dx
So, in this case we can make:
![Page 35: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/35.jpg)
Example 1
Let’s try to find the integral: ∫xexdx
This integral can’t be solved with any of our previous methods.So, let’s try integration by parts!Our formula is:∫
u(x)v ′(x)dx = u(x)v(x) −∫
u′(x)v(x)dx
So, in this case we can make:
u(x) = x
![Page 36: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/36.jpg)
Example 1
Let’s try to find the integral: ∫xexdx
This integral can’t be solved with any of our previous methods.So, let’s try integration by parts!Our formula is:∫
u(x)v ′(x)dx = u(x)v(x) −∫
u′(x)v(x)dx
So, in this case we can make:
u(x) = x v ′(x) = ex
![Page 37: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/37.jpg)
Example 1
Let’s try to find the integral: ∫xexdx
This integral can’t be solved with any of our previous methods.So, let’s try integration by parts!Our formula is:∫
u(x)v ′(x)dx = u(x)v(x) −∫
u′(x)v(x)dx
So, in this case we can make:
u(x) = x v ′(x) = ex
u′(x) = 1
![Page 38: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/38.jpg)
Example 1
Let’s try to find the integral: ∫xexdx
This integral can’t be solved with any of our previous methods.So, let’s try integration by parts!Our formula is:∫
u(x)v ′(x)dx = u(x)v(x) −∫
u′(x)v(x)dx
So, in this case we can make:
u(x) = x v ′(x) = ex
u′(x) = 1 v(x) = ex
![Page 39: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/39.jpg)
Example 1
Let’s try to find the integral: ∫xexdx
This integral can’t be solved with any of our previous methods.So, let’s try integration by parts!Our formula is:∫
u(x)v ′(x)dx = u(x)v(x) −∫
u′(x)v(x)dx
So, in this case we can make:
u(x) = x v ′(x) = ex
u′(x) = 1 v(x) = ex∫xexdx =
![Page 40: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/40.jpg)
Example 1
Let’s try to find the integral: ∫xexdx
This integral can’t be solved with any of our previous methods.So, let’s try integration by parts!Our formula is:∫
u(x)v ′(x)dx = u(x)v(x) −∫
u′(x)v(x)dx
So, in this case we can make:
u(x) = x v ′(x) = ex
u′(x) = 1 v(x) = ex
∫��u(x)
xexdx =
![Page 41: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/41.jpg)
Example 1
Let’s try to find the integral: ∫xexdx
This integral can’t be solved with any of our previous methods.So, let’s try integration by parts!Our formula is:∫
u(x)v ′(x)dx = u(x)v(x) −∫
u′(x)v(x)dx
So, in this case we can make:
u(x) = x v ′(x) = ex
u′(x) = 1 v(x) = ex
∫x��>
v ′(x)exdx =
![Page 42: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/42.jpg)
Example 1
Let’s try to find the integral: ∫xexdx
This integral can’t be solved with any of our previous methods.So, let’s try integration by parts!Our formula is:∫
u(x)v ′(x)dx = u(x)v(x) −∫
u′(x)v(x)dx
So, in this case we can make:
u(x) = x v ′(x) = ex
u′(x) = 1 v(x) = ex∫xexdx =
![Page 43: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/43.jpg)
Example 1
Let’s try to find the integral: ∫xexdx
This integral can’t be solved with any of our previous methods.So, let’s try integration by parts!Our formula is:∫
u(x)v ′(x)dx = u(x)v(x) −∫
u′(x)v(x)dx
So, in this case we can make:
u(x) = x v ′(x) = ex
u′(x) = 1 v(x) = ex∫xexdx = xex
![Page 44: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/44.jpg)
Example 1
Let’s try to find the integral: ∫xexdx
This integral can’t be solved with any of our previous methods.So, let’s try integration by parts!Our formula is:∫
u(x)v ′(x)dx = u(x)v(x) −∫
u′(x)v(x)dx
So, in this case we can make:
u(x) = x v ′(x) = ex
u′(x) = 1 v(x) = ex
∫xexdx =��
u(x)xex
![Page 45: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/45.jpg)
Example 1
Let’s try to find the integral: ∫xexdx
This integral can’t be solved with any of our previous methods.So, let’s try integration by parts!Our formula is:∫
u(x)v ′(x)dx = u(x)v(x) −∫
u′(x)v(x)dx
So, in this case we can make:
u(x) = x v ′(x) = ex
u′(x) = 1 v(x) = ex
∫xexdx = x��>
v(x)ex
![Page 46: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/46.jpg)
Example 1
Let’s try to find the integral: ∫xexdx
This integral can’t be solved with any of our previous methods.So, let’s try integration by parts!Our formula is:∫
u(x)v ′(x)dx = u(x)v(x) −∫
u′(x)v(x)dx
So, in this case we can make:
u(x) = x v ′(x) = ex
u′(x) = 1 v(x) = ex∫xexdx = xex−
![Page 47: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/47.jpg)
Example 1
Let’s try to find the integral: ∫xexdx
This integral can’t be solved with any of our previous methods.So, let’s try integration by parts!Our formula is:∫
u(x)v ′(x)dx = u(x)v(x) −∫
u′(x)v(x)dx
So, in this case we can make:
u(x) = x v ′(x) = ex
u′(x) = 1 v(x) = ex∫xexdx = xex −
∫1.exdx
![Page 48: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/48.jpg)
Example 1
Let’s try to find the integral: ∫xexdx
This integral can’t be solved with any of our previous methods.So, let’s try integration by parts!Our formula is:∫
u(x)v ′(x)dx = u(x)v(x) −∫
u′(x)v(x)dx
So, in this case we can make:
u(x) = x v ′(x) = ex
u′(x) = 1 v(x) = ex
∫xexdx = xex −
∫���
u′(x)
1.exdx
![Page 49: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/49.jpg)
Example 1
Let’s try to find the integral: ∫xexdx
This integral can’t be solved with any of our previous methods.So, let’s try integration by parts!Our formula is:∫
u(x)v ′(x)dx = u(x)v(x) −∫
u′(x)v(x)dx
So, in this case we can make:
u(x) = x v ′(x) = ex
u′(x) = 1 v(x) = ex
∫xexdx = xex −
∫1.��>
v(x)exdx
![Page 50: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/50.jpg)
Example 1
Let’s try to find the integral: ∫xexdx
This integral can’t be solved with any of our previous methods.So, let’s try integration by parts!Our formula is:∫
u(x)v ′(x)dx = u(x)v(x) −∫
u′(x)v(x)dx
So, in this case we can make:
u(x) = x v ′(x) = ex
u′(x) = 1 v(x) = ex∫xexdx = xex −
∫1.exdx
![Page 51: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/51.jpg)
Example 1
![Page 52: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/52.jpg)
Example 1
So, we only need to solve:
![Page 53: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/53.jpg)
Example 1
So, we only need to solve:∫xexdx = xex −
∫exdx
![Page 54: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/54.jpg)
Example 1
So, we only need to solve:∫xexdx = xex −
∫exdx
The integral in the second term is easy to solve:
![Page 55: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/55.jpg)
Example 1
So, we only need to solve:∫xexdx = xex −
∫exdx
The integral in the second term is easy to solve:∫xexdx = xex − ex
![Page 56: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/56.jpg)
Example 1
So, we only need to solve:∫xexdx = xex −
∫exdx
The integral in the second term is easy to solve:∫xexdx = xex − ex
∫xexdx = ex (x − 1)
![Page 57: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/57.jpg)
Example 1
So, we only need to solve:∫xexdx = xex −
∫exdx
The integral in the second term is easy to solve:∫xexdx = xex − ex
∫xexdx = ex (x − 1)
That’s the final answer.
![Page 58: Integration by Parts, Part 1](https://reader033.fdocuments.net/reader033/viewer/2022042510/54961c7ab479597e6a8b6193/html5/thumbnails/58.jpg)