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Evaluate each of the following indefinite integrals:
Evaluate $\int \left(4{x}^{3}-5{x}^{2}+x-7\right)dx}.$
${x}^{4}-\frac{5}{3}{x}^{3}+\frac{1}{2}{x}^{2}-7x+C$
We look at techniques for integrating a large variety of functions involving products, quotients, and compositions later in the text. Here we turn to one common use for antiderivatives that arises often in many applications: solving differential equations.
A differential equation is an equation that relates an unknown function and one or more of its derivatives. The equation
is a simple example of a differential equation. Solving this equation means finding a function $y$ with a derivative $f.$ Therefore, the solutions of [link] are the antiderivatives of $f.$ If $F$ is one antiderivative of $f,$ every function of the form $y=F\left(x\right)+C$ is a solution of that differential equation. For example, the solutions of
are given by
Sometimes we are interested in determining whether a particular solution curve passes through a certain point $\left({x}_{0},{y}_{0}\right)$ —that is, $y\left({x}_{0}\right)={y}_{0}.$ The problem of finding a function $y$ that satisfies a differential equation
with the additional condition
is an example of an initial-value problem . The condition $y\left({x}_{0}\right)={y}_{0}$ is known as an initial condition . For example, looking for a function $y$ that satisfies the differential equation
and the initial condition
is an example of an initial-value problem. Since the solutions of the differential equation are $y=2{x}^{3}+C,$ to find a function $y$ that also satisfies the initial condition, we need to find $C$ such that $y\left(1\right)=2{\left(1\right)}^{3}+C=5.$ From this equation, we see that $C=3,$ and we conclude that $y=2{x}^{3}+3$ is the solution of this initial-value problem as shown in the following graph.
Solve the initial-value problem
First we need to solve the differential equation. If $\frac{dy}{dx}=\text{sin}\phantom{\rule{0.1em}{0ex}}x,$ then
Next we need to look for a solution $y$ that satisfies the initial condition. The initial condition $y\left(0\right)=5$ means we need a constant $C$ such that $\text{\u2212}\text{cos}\phantom{\rule{0.1em}{0ex}}x+C=5.$ Therefore,
The solution of the initial-value problem is $y=\text{\u2212}\text{cos}\phantom{\rule{0.1em}{0ex}}x+6.$
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