∫kf(u) du
To integrate the function ∫kf(u) du, where “k” is a constant and “f(u)” is a function of “u”, we use the constant multiple rule of integration:
∫kf(u) du = k ∫f(u) du
This rule states that we can pull out the constant “k” outside the integral sign
To integrate the function ∫kf(u) du, where “k” is a constant and “f(u)” is a function of “u”, we use the constant multiple rule of integration:
∫kf(u) du = k ∫f(u) du
This rule states that we can pull out the constant “k” outside the integral sign.
Now, the integral of f(u) du can be evaluated based on the specific function f(u). The technique for integrating f(u) depends on the form of the function.
Let’s consider a few examples:
1. ∫k du (where f(u) = 1)
In this case, f(u) is a constant as well. So, when we integrate ∫f(u) du, we get f(u)u. Therefore,
∫k du = k ∫1 du = k(u) + C,
where C is the constant of integration.
2. ∫k√u du (where f(u) = √u)
Here, f(u) = √u is a function involving u. So, we can integrate ∫f(u) du by applying the power rule for integration.
∫k√u du = k * [(2/3) * u^(3/2)] + C
Simplifying further, we have
∫k√u du = (2k/3) * u^(3/2) + C.
Again, C represents the constant of integration.
These are just a couple of examples, and the method of integration will vary depending on the specific function f(u) involved. However, the general process involves using the constant multiple rule and then applying the appropriate integration techniques based on the form of f(u).
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