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Title: Second Lecture of Calculus II for Physical Sciences
Description: Summary of the contents, with examples, of the second lecture of Calculus II for Physical Sciences at the University of Toronto Scarborough.
Description: Summary of the contents, with examples, of the second lecture of Calculus II for Physical Sciences at the University of Toronto Scarborough.
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University of Toronto Scarborough
MATA36
Calculus II for Physical Sciences
Text Book: Calculus Early Transcendentals 9th Edition by James
Stewart, Daniel Clegg, Saleem Watson
In calculus, there are two widely used methods of integrating, integration by substitution and
integration by parts
...
Here, we seek to substitute a function inside a function, within the function we are provided to
integrate, with the variable u, then we find du, i
...
, find the derivative of u in terms of x, then
rewrite the equation in terms of dx, and substitute the value of dx in the original function, and
proceed onwards
...
Then, letβs find the derivative of the substitution
...
Letβs rearrange the equation with dx on the left-hand side
...
Now, letβs substitute the value of dx into the original function
...
Simplifying, we get,
β« π’10 ππ’
Now, this is an integral we can easily solve using the formula β« π₯ π ππ₯ =
π₯ π+1
π+1
ππ₯
...
π’11
11
+πΆ =
(1+π₯ 3 )11
11
+ πΆ (This is the final answer)
...
Integration by parts
it is the reverse or antiderivative version of the product rule for calculating derivatives
...
Letβs elucidate with an example
...
Whatever notation you use, the meaning is same, you
consider the function as a product of two functions, then derive one, integrate the other, and
apply the above formula
...
The
goal is to endure that the integration on the right-hand side of the function, i
...
, β« π£ππ’ or
β« π(π₯)πβ²(π₯) is a simpler integration than the original function which we are provided to
integrate
...
Let x = u and e-x = dv
...
Now simplifying, we have,
= π₯(βπ βπ₯ ) β β«(βπ βπ₯ )(1) ππ₯
= βπ₯π βπ₯ β (β β« π βπ₯ ππ₯)
= βπ₯π βπ₯ + β« π βπ₯ ππ₯
= βπ₯π βπ₯ + (βπ βπ₯ )
= βπ₯π βπ₯ β π βπ₯ + πΆ (This is the final answer)
Now, what would happen if we try to switch the notations? Letβs seeβ¦
...
Then, du = -e-x and π£ = β« π₯ ππ₯ =
π₯2
2
Now, applying the above formula, we have,
β« π₯π βπ₯ = (π βπ₯ ) (
π₯2
π₯2
) β β« ( ) (βπ βπ₯ ) ππ₯
2
2
π₯2
Now, we can see that the new integration we haven the right-hand side, β β« ( ) (βπ βπ₯ ) ππ₯
2
β, is more complex than the original integration we had, β β« π₯π
that the notations need to be switched
...
This is a red flag, signaling
Title: Second Lecture of Calculus II for Physical Sciences
Description: Summary of the contents, with examples, of the second lecture of Calculus II for Physical Sciences at the University of Toronto Scarborough.
Description: Summary of the contents, with examples, of the second lecture of Calculus II for Physical Sciences at the University of Toronto Scarborough.