UpStudy Free Solution:

To solve these definite integrals, let's first understand the given information:

\(\int _ { 0} ^ { 2} f( z) \space dz = 3\)

\(\int _ { 0} ^ { 3} f( z) \space dz = 8\)
(a) \(\int _ { 2} ^ { 3} f( z) \space dz = 5\)
We can use the property of definite integrals that states:
\[\int _ { a} ^ { b} f( z) \space dz + \int _ { b} ^ { c} f( z) \space dz = \int _ { a} ^ { c} f( z) \space dz\]
In this case, \(a = 0\), \(b = 2\), and \(c = 3\):
\[\int _ { 0} ^ { 2} f( z) \space dz + \int _ { 2} ^ { 3} f( z) \space dz = \int _ { 0} ^ { 3} f( z) \space dz\]
Substituting the given values:
\[3 + \int _ { 2} ^ { 3} f( z) \space dz = 8\]
Solving for \(\int _ { 2} ^ { 3} f( z) \space dz\):
\[\int _ { 2} ^ { 3} f( z) \space dz = 8 - 3 = 5\]
So, the answer for (a) is:
\[\int _ { 2} ^ { 3} f( z) \space dz = 5\]
(b) \(\int _ { 3} ^ { 3} f( z) \space dz = \square \)
The integral of any function over an interval where the upper and lower limits are the same is always 0:
\[\int _ { a} ^ { a} f( z) \space dz = 0\]
So, for \(\int _ { 3} ^ { 3} f( z) \space dz\):
\[\int _ { 3} ^ { 3} f( z) \space dz = 0\]
So, the answer for (b) is:
\[0\]
Summary
(a) \(\int _ { 2} ^ { 3} f( z) \space dz = 5\)
(b) \(\int _ { 3} ^ { 3} f( z) \space dz = 0\)

Supplemental Knowledge

A definite integral represents the area under the curve of a function \(f( x) \) between two points \(a\) and \(b\). The notation for a definite integral is:
\[\int _ { a} ^ { b} f( x) \space dx\]
Given that the function \(f\) is integrable, we can use properties of definite integrals to solve problems. Two important properties are:

Additivity over Intervals:
\[\int _ { a} ^ { c} f( x) \space dx + \int _ { c} ^ { b} f( x) \space dx = \int _ { a} ^ { b} f( x) \space dx\]

Zero Interval Property:
If the limits of integration are the same, then:
\[\int _ { a} ^ { a} f( x) \space dx = 0\]
These properties help in evaluating integrals over different intervals and understanding their behavior.
 

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