91Ó°ÊÓ

Integration by substitution is a method used to simplify an integral by making a substitution for a part of the integrand. This involves selecting a new variable to replace the more complicated expression. For instance, given the integral \ \[ \int_{-9}^{-1} \frac{y dy}{\sqrt{4-5y}} \ \], we can use substitution to handle the expression under the square root. First, we let \ u = 4 - 5y \, and then differentiate both sides to find \ du = -5 \ dy \. Hence, \ dy = -\frac{1}{5} \ du \. By performing this substitution, the complex integral is transformed into a simpler one that may be easier to evaluate. Remember to also convert the limits of integration: when \ y = -9 \, then \ u = 49 \, and when \ y = -1 \, then \ u = 9 \. This method significantly eases the evaluation of the integral.

In calculus, integrals containing square roots can often be tricky. The presence of a square root, especially in the integrand, can complicate the integration process. For example, in the integral \ \( \int_{49}^{9} \frac{(\frac{4-u}{5})(-\frac{1}{5} du)}{\sqrt{u}} \)\, the square root symbol \( \sqrt{u} \) affects how we proceed. Using our substitution of \ u = 4 - 5y \, the term under the square root simplifies into a form that gives us \ u \ as our new variable for the limits of integration. We then substitute \ y \, \ dy \, and the limits, re-writing the integral in terms of \ u \ where we eventually get \ \( -\frac{1}{25} \int_{49}^{9} \frac{4 - u}{\frac{1}{2}} du \)\. This separation into manageable parts is very helpful when dealing with square root integrals. Once simplified, each part can be integrated separately to obtain the final result.

A definite integral involves integrating over a specified interval, from one limit to another, providing the area under a curve between these two points. In our example \ \(\frac{y dy}{\frac{(4-5y)}}{\sqrt{4-5y}}\)\, substituting \ y = -9 \, we find \ u = 49 \ and for \ y = -1 \, we find \ u = 9 \. This substitution transforms the original definite integral into \ \ \( \int_{49}^{9} -\frac{1}{25} \frac{4 - u}{\frac{u \sqrt{u}\}} du \). The integral is then solved within these new limits. Definite integrals provide a numerical value representing the net area under the curve between two points, which in this case is calculated as \ \(\frac{728}{75}\)\ after evaluating and combining the simplified integral parts, allowing us to understand the total accumulation of quantities described by the integrand between specified limits.