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Evaluating indefinite integrals

Evaluate each of the following indefinite integrals:

  1. ( 5 x 3 7 x 2 + 3 x + 4 ) d x
  2. x 2 + 4 x 3 x d x
  3. 4 1 + x 2 d x
  4. tan x cos x d x
  1. Using [link] , we can integrate each of the four terms in the integrand separately. We obtain
    ( 5 x 3 7 x 2 + 3 x + 4 ) d x = 5 x 3 d x 7 x 2 d x + 3 x d x + 4 d x .

    From the second part of [link] , each coefficient can be written in front of the integral sign, which gives
    5 x 3 d x 7 x 2 d x + 3 x d x + 4 d x = 5 x 3 d x 7 x 2 d x + 3 x d x + 4 1 d x .
    Using the power rule for integrals, we conclude that
    ( 5 x 3 7 x 2 + 3 x + 4 ) d x = 5 4 x 4 7 3 x 3 + 3 2 x 2 + 4 x + C .
  2. Rewrite the integrand as
    x 2 + 4 x 3 x = x 2 x + 4 x 3 x = 0 .

    Then, to evaluate the integral, integrate each of these terms separately. Using the power rule, we have
    ( x + 4 x 2 / 3 ) d x = x d x + 4 x −2 / 3 d x = 1 2 x 2 + 4 1 ( −2 3 ) + 1 x ( −2 / 3 ) + 1 + C = 1 2 x 2 + 12 x 1 / 3 + C .
  3. Using [link] , write the integral as
    4 1 1 + x 2 d x .

    Then, use the fact that tan −1 ( x ) is an antiderivative of 1 ( 1 + x 2 ) to conclude that
    4 1 + x 2 d x = 4 tan −1 ( x ) + C .
  4. Rewrite the integrand as
    tan x cos x = sin x cos x cos x = sin x .

    Therefore,
    tan x cos x = sin x = cos x + C .
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Evaluate ( 4 x 3 5 x 2 + x 7 ) d x .

x 4 5 3 x 3 + 1 2 x 2 7 x + C

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Initial-value problems

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

d y d x = f ( x )

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 ( x ) + C is a solution of that differential equation. For example, the solutions of

d y d x = 6 x 2

are given by

y = 6 x 2 d x = 2 x 3 + C .

Sometimes we are interested in determining whether a particular solution curve passes through a certain point ( x 0 , y 0 ) —that is, y ( x 0 ) = y 0 . The problem of finding a function y that satisfies a differential equation

d y d x = f ( x )

with the additional condition

y ( x 0 ) = y 0

is an example of an initial-value problem . The condition y ( x 0 ) = y 0 is known as an initial condition . For example, looking for a function y that satisfies the differential equation

d y d x = 6 x 2

and the initial condition

y ( 1 ) = 5

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 ( 1 ) = 2 ( 1 ) 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.

The graphs for y = 2x3 + 6, y = 2x3 + 3, y = 2x3, and y = 2x3 − 3 are shown.
Some of the solution curves of the differential equation d y d x = 6 x 2 are displayed. The function y = 2 x 3 + 3 satisfies the differential equation and the initial condition y ( 1 ) = 5 .

Solving an initial-value problem

Solve the initial-value problem

d y d x = sin x , y ( 0 ) = 5 .

First we need to solve the differential equation. If d y d x = sin x , then

y = sin ( x ) d x = cos x + C .

Next we need to look for a solution y that satisfies the initial condition. The initial condition y ( 0 ) = 5 means we need a constant C such that cos x + C = 5 . Therefore,

C = 5 + cos ( 0 ) = 6 .

The solution of the initial-value problem is y = cos x + 6 .

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Source:  OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2
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