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Question
(description: sketch of smooth continuous curve, below x - axis between x = 0 and x = 3, shaded and labeled 10; above x - axis between x = 3 and x = 6, shaded and labeled 8; below x - axis between x = 6 and x = 8, shaded and labeled 5.)
- 5 pts, 1 each above is the graph of a function ( f(x) ) with the numbers in the shaded regions indicating their areas. compute the following:
- ( int_{0}^{3}f(x)dx )
- ( int_{3}^{6}f(x)dx )
- ( int_{0}^{8}|f(x)|dx )
- ( int_{0}^{8}(-f(x))dx )
- ( int_{8}^{0}f(x)dx )
Step1: Recall the property of definite integral and area
If \(y = f(x)\) is below the \(x -\)axis on \([a,b]\), then \(\int_{a}^{b}f(x)dx=-\text{(Area between }y = f(x)\text{ and }x-\text{axis)}\). If \(y = f(x)\) is above the \(x -\)axis on \([a,b]\), then \(\int_{a}^{b}f(x)dx=\text{(Area between }y = f(x)\text{ and }x-\text{axis)}\)
Step2: Calculate \(\int_{0}^{3}f(x)dx\)
Since the function \(y = f(x)\) is below the \(x -\)axis on \([0,3]\), \(\int_{0}^{3}f(x)dx=- 10\)
Step3: Calculate \(\int_{3}^{6}f(x)dx\)
Since the function \(y = f(x)\) is above the \(x -\)axis on \([3,6]\), \(\int_{3}^{6}f(x)dx = 8\)
Step4: Calculate \(\int_{0}^{8}|f(x)|dx\)
\(\int_{0}^{8}|f(x)|dx=\int_{0}^{3}|f(x)|dx+\int_{3}^{6}|f(x)|dx+\int_{6}^{8}|f(x)|dx\). Since \(|f(x)|=-f(x)\) on \([0,3]\) and \([6,8]\) (where \(f(x)<0\)) and \(|f(x)| = f(x)\) on \([3,6]\) (where \(f(x)>0\)), \(\int_{0}^{8}|f(x)|dx=10 + 8+5=23\)
Step5: Calculate \(\int_{0}^{8}(-f(x))dx\)
\(\int_{0}^{8}(-f(x))dx=-\int_{0}^{8}f(x)dx\). We know that \(\int_{0}^{8}f(x)dx=\int_{0}^{3}f(x)dx+\int_{3}^{6}f(x)dx+\int_{6}^{8}f(x)dx=-10 + 8-5=-7\). So \(\int_{0}^{8}(-f(x))dx=-(-7) = 7\)
Step6: Calculate \(\int_{8}^{0}f(x)dx\)
Using the property \(\int_{a}^{b}f(x)dx=-\int_{b}^{a}f(x)dx\), \(\int_{8}^{0}f(x)dx=-\int_{0}^{8}f(x)dx=-(-7) = 7\)
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\(\int_{0}^{3}f(x)dx=-10\)
\(\int_{3}^{6}f(x)dx = 8\)
\(\int_{0}^{8}|f(x)|dx=23\)
\(\int_{0}^{8}(-f(x))dx = 7\)
\(\int_{8}^{0}f(x)dx=7\)