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integration
- Thread starter maney21
- Start date
BigGlenntheHeavy
Senior Member
- Joined
- Mar 8, 2009
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- 1,577
\(\displaystyle \int_{0}^{1}20sin\bigg(\frac{\pi t}{30}\bigg)dt \ = \ 20\int_{0}^{1}sin\bigg(\frac{\pi t}{30}\bigg)dt\)
\(\displaystyle Now \ let \ u \ = \ \frac{\pi t}{30}, \ du \ = \ \frac{\pi dt}{30}, \ \frac{30du}{\pi} \ = \ dt\)
\(\displaystyle Ergo, \ we \ have: \ 20\bigg(\frac{30}{\pi}\bigg)\int_{0}^{\pi/30}sin(u)du \ = \ \frac{600}{\pi}\bigg[-cos(u)\bigg]_{0}^{\pi/30}\)
\(\displaystyle = \ \frac{600}{\pi}[1-cos(\pi/30)] \ = \ 1.0462409171\)
\(\displaystyle Now \ let \ u \ = \ \frac{\pi t}{30}, \ du \ = \ \frac{\pi dt}{30}, \ \frac{30du}{\pi} \ = \ dt\)
\(\displaystyle Ergo, \ we \ have: \ 20\bigg(\frac{30}{\pi}\bigg)\int_{0}^{\pi/30}sin(u)du \ = \ \frac{600}{\pi}\bigg[-cos(u)\bigg]_{0}^{\pi/30}\)
\(\displaystyle = \ \frac{600}{\pi}[1-cos(\pi/30)] \ = \ 1.0462409171\)