5.3 The Fundamental Theorem of Calculus/11: Difference between revisions
Jump to navigation
Jump to search
No edit summary |
m (Protected "5.3 The Fundamental Theorem of Calculus/11" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))) |
||
| (One intermediate revision by the same user not shown) | |||
| Line 1: | Line 1: | ||
<math> g(x)= \int_{x}^{\pi}\sqrt{1+sec(t)}\cdot dt </math> <br><br> | <math> g(x)= \int_{x}^{\pi}\sqrt{1+sec(t)}\cdot dt </math> <br><br> | ||
<math> \frac{d}{dx}\left[g(x)\right]=\frac{d}{dx}\left[\ | <math> \frac{d}{dx}\left[g(x)\right]=\frac{d}{dx}\left[\int_x^{\pi}\sqrt{1+sec(t)}\cdot dt\right]=0 \cdot \sqrt{1+sec(\pi)} - 1\cdot \sqrt{1+sec(x)} = -\sqrt{1+sec(x)}</math> <br><br> | ||
<math>\text{Therefore, } g'(x)=-\sqrt{1+sec(x)}</math> | <math>\text{Therefore, } g'(x)=-\sqrt{1+sec(x)}</math> | ||
Latest revision as of 20:15, 6 September 2022
![{\displaystyle {\frac {d}{dx}}\left[g(x)\right]={\frac {d}{dx}}\left[\int _{x}^{\pi }{\sqrt {1+sec(t)}}\cdot dt\right]=0\cdot {\sqrt {1+sec(\pi )}}-1\cdot {\sqrt {1+sec(x)}}=-{\sqrt {1+sec(x)}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/61cd7fe7ef01b603b4b8f95f5036e31f26c4e84a)
