Computing an integral by substitution is the reverse of the chain rule for computing the derivative. Substitution is intended to rewrite a complicated-looking integral involving the derivative of some component expression as another much simpler integral. For example, if $u=f(x)$ , then $\mathrm du=f'(x)\,\mathrm dx$ and

$\displaystyle\int f(x)f'(x)\,\mathrm dx=\int u\,\mathrm du$

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Integration by parts is the reverse of the product rule for derivatives:

$(f(x)g(x))'=f'(x)g(x)+f(x)g'(x)\implies f(x)g'(x)=(f(x)g(x))'-f'(x)g(x)$

Integrating both sides with respect to $x$ gives

$\displaystyle\int f(x)g'(x)\,\mathrm dx=\int(f(x)g(x))'\,\mathrm dx-\int f'(x)g(x)\,\mathrm dx$

$\displaystyle\int f(x)g'(x)\,\mathrm dx=f(x)g(x)-\int f'(x)g(x)\,\mathrm dx$

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Personally, I think the best way to grasp the idea behind the two methods is to practice. You start to notice patterns to the point where knowing which is the "right" method to use becomes second nature.

4 0

3 years ago

Suppose f(x)=x find the graph of f(x)+1

emmasim [6.3K]

Graph of f(x)+1 would be x+1.
This would mean that all y-values of the graph will increase by one.

7 0

3 years ago

I need to know if these rates are equivalent

8090 [49]

No these rate are not equivalent hope this helped

7 0

4 years ago