Notes
| \({\text{Properties of Integrals}}\) |
| \(\underline{\text{Name}}\) |
\(\underline{\text{Formula}}\) |
| \(\text{Sum & Difference}\) |
\(\displaystyle \int_{a}^{b}\left(f(x) \pm g(x)\right)\,dx \,=\, \int_{a}^{b}f(x)\,dx \pm \int_{a}^{b}g(x)\,dx\) |
| \(\text{Constant Multiple}\) |
\(\displaystyle \int_{a}^{b}\left(k \cdot f(x) \right)\,dx \,=\, k\cdot \int_{a}^{b}f(x)\,dx \) |
| \(\text{Order of Interval}\) |
\(\displaystyle \int_{a}^{b}f(x) \,dx \,=\, – \int_{b}^{a}f(x)\,dx \) |
| \(\text{Zero}\) |
\(\displaystyle \int_{a}^{a}f(x)\,dx \,=\, 0\) |
| \(\text{Adding Intervals}\) |
\(\displaystyle \int_{a}^{b}f(x)\,dx + \int_{b}^{c}f(x)\,dx \,=\, \int_{a}^{c}f(x)\,dx\) |
Questions & Solutions
\(\textbf{1)}\) \( \displaystyle \int_{2}^{2}f(x)\,dx \)
The answer is \(0 \)
\(\,\,\,\,\,\displaystyle \int_{a}^{a}f(x)\,dx \,=\, 0\)
\(\textbf{2)}\) \( \displaystyle \int_{2}^{9}3\,dx \)
\(3(9-2)=21 \)
\(\textbf{3)}\) \( \displaystyle \int_{8}^{8}f(x)\,dx \)
The answer is \(0 \)
\(\textbf{4)}\) \( \displaystyle \int_{1}^{10}5\,dx \)
\(5(10-1)=45 \)
\(\textbf{5)}\) \(\displaystyle \int_{2}^{5}f(x)\,dx=7, \enspace \displaystyle \int_{5}^{9}f(x)\,dx=4, \enspace \displaystyle \int_{2}^{12}f(x)dx=16 \)
\( \text{Solve for } \displaystyle \int_{2}^{9}f(x)\,dx \)
The answer is \( 11 \)
\(\text{Notes}\)
\(\,\,\,\,\,\displaystyle\, \int_{a}^{c}f(x)\,dx \, = \, \int_{a}^{b}f(x)\,dx + \int_{b}^{c}f(x)\,dx\)
\(\text{Apply notes to this problem.}\)
\(\,\,\,\,\,\displaystyle \int_{2}^{9} f(x)\,dx \,= \,f(x)dx\displaystyle \int_{2}^{5}f(x)\,dx+\displaystyle \int_{5}^{9}f(x)\,dx \)
\(\,\,\,\,\,\displaystyle \int_{2}^{9} f(x)\,dx \,= 7+4 \)
\(\,\,\,\,\,\displaystyle \int_{2}^{9} f(x)\,dx \,= 11 \)
\(\textbf{6)}\) \(\displaystyle \int_{2}^{5}f(x)\,dx=7, \enspace \displaystyle \int_{5}^{9}f(x)\,dx=4, \enspace \displaystyle \int_{2}^{12}f(x)\,dx=16 \)
\( \text{Solve for } \displaystyle \int_{5}^{12}f(x)\,dx \)
\(\displaystyle \int_{2}^{12}f(x)\,dx-\displaystyle \int_{2}^{5}f(x)\,dx=16-7=9 \)
\(\textbf{7)}\) \(\displaystyle \int_{2}^{5}f(x)\,dx=7, \enspace \displaystyle \int_{5}^{9}f(x)\,dx=4, \enspace \displaystyle \int_{2}^{12}f(x)\,dx=16 \)
\( \text{Solve for } \displaystyle \int_{9}^{5}f(x)\,dx \)
\(-\displaystyle \int_{5}^{9}f(x)\,dx=-4 \)
\(\textbf{8)}\) \(\displaystyle \int_{2}^{5}f(x)\,dx=7, \enspace \displaystyle \int_{5}^{9}f(x)\,dx=4, \enspace \displaystyle \int_{2}^{12}f(x)\,dx=16 \)
\( \text{Solve for } \displaystyle \int_{2}^{5}4f(x)\,dx \)
\(4\displaystyle \int_{2}^{5}f(x)\,dx=4\cdot7=28\)
\(\textbf{9)}\) If\( \displaystyle \int_{a}^{b}f(x)\,dx=5a-2b \), then what is \( \displaystyle \int_{a}^{b}[f(x)-3]dx? \)
\(5a-2b-3(b-a)=8a-5b \)
\(\text{Notes}\)
\(\,\,\,\,\,\displaystyle\, \int_{a}^{c}f(x)\,dx \, = \, \int_{a}^{b}f(x)\,dx + \int_{b}^{c}f(x)\,dx\)
\(\text{Apply notes to this problem.}\)
\(\,\,\,\,\,\displaystyle \int_{a}^{b} [f(x)-3]\,dx\)
\(\,\,\,\,\,\displaystyle \int_{a}^{b}f(x)\,dx – \displaystyle \int_{a}^{b}3\,dx \)
\(\,\,\,\,\,\displaystyle \int_{a}^{b}f(x)\,dx – 3 \displaystyle \int_{a}^{b}1\,dx \)
\(\,\,\,\,\,\displaystyle 5a-2b – 3(b-a) \)
\(\,\,\,\,\,\displaystyle 5a-2b – 3b +3a \)
\(\,\,\,\,\,\displaystyle 8a-5b \)
See Related Pages\(\)