# U-substitution (also called change of variable) is a way to undo: a) Rotations across the y-axis.

## Question:

U-substitution (also called change of variable) is a way to undo:

a) Rotations across the y-axis. |

b) The chain rule. |

c) Complicated rates of change. |

d) Global warming. |

## U-Substitution as an Integration Technique

When we evaluate an indefinite integral, we need to find the antiderivative of an integrand. This is sometimes a difficult procedure, but performing substitution can allow us to rewrite a complex integrand as one more simple. This may allow us to find the antiderivative we're looking for.

## Answer and Explanation:

b) The Chain Rule

When we perform u-substitution, we're applying a technique to evaluate an indefinite integral. This means we're trying to find an antiderivative for a specific integrand. As the antiderivative is the opposite of a derivative, we're finding a way to undo a differentiation technique. The Chain Rule is one differentiation technique that can be addressed, specifically by using u-substitution as stated.

Recall that the Chain Rule is used to differentiate a composition of functions, {eq}(f \circ g )(x) {/eq}. As an integrand is the result of a hypothetical differentiation problem, where the function being differentiated is the result of integration, this derivative must have been found by applying the Chain Rule. By performing the following substitution, we can find the solution to this integral.

One example of an integration problem that involves substitution is below. Note that we need to assume that {eq}g'(x) {/eq} lies within this integrand, as we can't apply this technique otherwise.

{eq}\displaystyle \int f(g(x)) \cdot g'(x) dx {/eq}

If we perform u-substitution, letting {eq}u = g(x) {/eq} and {eq}du = g'(x) dx {/eq}, then we can simplify this as follows.

{eq}\displaystyle \int f(g(x)) \cdot g'(x) dx = \int f(u) du {/eq}

Let's let F be the antiderivative of f, which allows us to find the solution below.

{eq}\begin{align*} \displaystyle \int f(g(x)) \cdot g'(x) dx &= \int f(u) du\\ &= F(u) + c\\ &= F(g(x)) + c \end{align*} {/eq}

Note that differentiating this result would require us to apply the Chain Rule, which supports our claim that this technique is a way to undo the Chain Rule.

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