Compute the following integral:
Since on the integral and on we have,
Compute the following integral:
Since on the integral and on we have,
Let
Find the area between the graphs of and on the interval .
First, we draw the graph, shading the region between the two graphs in blue.
In this problem on the entire interval ; however, the function is defined piecewise on this interval.
The functions and are defined piecewise on the interval by
and
So, we break each integral into two pieces and compute:
Let
Find the area between the graphs of and on the interval .
First, we draw the graph, shading the region between the two graphs in blue.
In this problem on the entire interval ; however, the function is defined piecewise on this interval as
Thus, we break the integral into intervals and , and in both cases we are taking the integral of , but the definition of will depend on which interval we are in. The computation is as follows:
Let
Find the area between the graphs of and on the interval .
First, we draw the graph, shading the region between the two graphs in blue.
In this problem on the entire interval ; however, the function is defined piecewise on this interval as
Thus, we break the integral into intervals and , and in both cases we are taking the integral of , but the definition of will depend on which interval we are in. The computation is as follows:
Let
Find the area between the graphs of and on the interval .
First, we draw the graph, shading the region between the two graphs in blue.
In this problem on the entire interval ; however, the function is defined piecewise on this interval as
Thus, we break the integral into intervals and , and in both cases we are taking the integral of , but the definition of will depend on which interval we are in. The computation is as follows:
Find all real numbers such that the following equations hold:
Solving for we have,
So, for the equation to hold we must have
Since the expression will be nonzero for any , we must have .
Compute
Since we are taking the integral of an absolute value, we want to break the expression in the absolute value into intervals on which it is either always positive or always negative, and evaluate the integrals of the pieces separately. So, examining we have zeros of at . If the expression is going to change sign it must do so at these points. So,
So, we break the integral into pieces (which we can do since the integral is additive with respect to the interval of integration by Theorem 1.17 of Apostol),
Now we can use the formula for the integral of a polynomial to calculate each of these separately,
Putting all of these pieces together, we have,
Define for all . Compute:
Decide whether the following are true or false.
Hence,
This means .
Hence, in either case and so .
Write the following inequalities in equivalent forms without the absolute values.
a1 is equivalent to b2:
a2 is equivalent to b5:
a3 is equivalent to b7:
a4 is equivalent to b10:
a5 is equivalent to b3:
a6 is equivalent to b8:
a7 is equivalent to b9:
a8 is equivalent to b4:
a9 is equivalent to b6:
a10 is equivalent to b1: and . Thus, we must have and . The first of these inequalities requires and to both be positive or both be negative. Thus, or . The second requires and to have opposite signs. So, . Combining these restrictions on we have (b1).