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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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Chapters

0:00Read signed area3:00Cancellation3:30Add adjacent intervals

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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: 1.5−1.5=01.5-1.5=0. The integral measures net signed area.

Splitting at the graph's breakpoints gives ∫07f(x) dx=4.5+6+0=10.5\int_0^7 f(x)\,dx=4.5+6+0=10.5. Additivity lets simpler geometric regions solve the larger interval.

Knowledge cards

01

Read signed area

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.

02

Cancellation

On the interval from 5 to 7, equal triangular areas above and below the axis cancel: 1.5−1.5=01.5-1.5=0. The integral measures net signed area.

03

Add adjacent intervals

Splitting at the graph's breakpoints gives ∫07f(x) dx=4.5+6+0=10.5\int_0^7 f(x)\,dx=4.5+6+0=10.5. Additivity lets simpler geometric regions solve the larger interval.

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  • Definite integrals ExplanationAt 0:00
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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: 1.5−1.5=01.5-1.5=0. The integral measures net signed area. Splitting at the graph's breakpoints gives ∫07f(x) dx=4.5+6+0=10.5\int_0^7 f(x)\,dx=4.5+6+0=10.5. Additivity lets simpler geometric regions solve the larger interval.

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