5.3 The Fundamental Theorem of Calculus/31: Difference between revisions

Latest revision as of 21:25, 6 September 2022

$\int _{0}^{\frac {\pi }{4}}\sec ^{2}(t)\,dt=\tan(t){\bigg |}_{0}^{\frac {\pi }{4}}=\tan \left({\frac {\pi }{4}}\right)-\tan(0)=1-0=1$

@@ Line 1: / Line 1: @@
-<math>\int_{0}^{\frac{\pi\}{4}}\sec^2(t)dt
+<math>
-=tan(\frac{\pi\}{4})-tan(0)
+\int_{0}^{\frac{\pi}{4}}\sec^{2}(t)\,dt = \tan(t)\bigg|_{0}^{\frac{\pi}{4}}=\tan\left(\frac{\pi}{4}\right)-\tan(0)=1-0=1
-=1-0 = 1</math>
+</math>