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A region above the horizontal axis contributes positively to the definite integral; a region below it contributes negatively. The first triangular region has integral 4.5.
Calculus / English
Evaluate Definite Integrals from a Graph Using Area
Mathispower4u · YouTube · 5:30
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Evaluate definite integrals from a piecewise-linear graph using signed geometric areas and interval additivity.
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0:00Read signed area3:00Cancellation3:30Add adjacent intervalsLearning script
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A region above the horizontal axis contributes positively to the definite integral; a region below it contributes negatively. The first triangular region has integral 4.5.
On the interval from 5 to 7, equal triangular areas above and below the axis cancel: . The integral measures net signed area.
Splitting at the graph's breakpoints gives . Additivity lets simpler geometric regions solve the larger interval.
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Cancellation
On the interval from 5 to 7, equal triangular areas above and below the axis cancel: . The integral measures net signed area.
Add adjacent intervals
Splitting at the graph's breakpoints gives . Additivity lets simpler geometric regions solve the larger interval.
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- Definite integrals ExplanationAt 0:00
Why this connection?
A region above the horizontal axis contributes positively to the definite integral; a region below it contributes negatively. The first triangular region has integral 4.5. On the interval from 5 to 7, equal triangular areas above and below the axis cancel: . The integral measures net signed area. Splitting at the graph's breakpoints gives . Additivity lets simpler geometric regions solve the larger interval.
Questions this video answers
Find a method↗
You use interval additivity by splitting the total interval at the graph's breakpoints into simpler sub-intervals. You calculate the signed area for each sub-interval separately and then sum these values to find the total definite integral.
Conditions: The function is piecewise-linear.; The graph has identifiable breakpoints where the slope changes or crosses the axis.
3:30 – 5:30Watch this moment ↗
Meet the concept↗
The value of the definite integral from 0 to 7 is 10.5. This is obtained by summing the signed areas of the three sub-intervals created by splitting at the breakpoints: 4.5, 6, and 0.
Conditions: The function is piecewise-linear.; The integration interval is from 0 to 7.; The graph consists of regions with signed areas 4.5, 6, and 0.
3:30 – 5:30Watch this moment ↗
Find a method↗
A region above the horizontal axis contributes positively to the definite integral, while a region below it contributes negatively. The definite integral measures net signed area, not total geometric area.
Conditions: The function is piecewise-linear or its graph encloses simple geometric regions.; The integral is evaluated over an interval where the graph crosses the horizontal axis.
0:00 – 3:00Watch this moment ↗
Understand why↗
Equal triangular areas above and below the axis result in a zero definite integral because the integral measures net signed area. The positive contribution from the region above the axis cancels out the negative contribution from the region below the axis.
Conditions: The interval contains regions both above and below the horizontal axis.; The geometric areas of these regions are equal.
3:00 – 3:30Watch this moment ↗
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