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BC 1-2 FTC Name: The (First) Fundamental Theorem of Calculus Let ƒ be a continuous function defined in an open interval I containing a. ò Then F(x) = x f (t)dt is a differentiable function for all x in I and x d f (t) dt = f (x) . dx òa ( a ) Find F(x) = ò (t 4 + 2t + 3) dt . x (1) 3 Then differentiate to find (2) Find ò x -2 (4) Find Find 4 3 d dx ( ò (sin 2t + cos3t) dt) . x -2 dæ x sin(t 2 )dt öø . In this case the ò è 17 dx 1 x3 sec 2 t dt . Then find IMSA (ò ( t + 2t + 3) dt). x (sin 2t + cos3t ) dt . Then find (3) dF d = dx dx d x 3 2 sec t dt dx 1 Euler5.1 S 12 BC 1-2 FTC Name: Note that problem (4) came out differently, due to the x3 as the upper limit of integration. The chain rule comes into play as we take the derivative. Continue with the next problems. (5) Find t sin x 3 2 Then find 5t 2 dt . d dx sin x 2 t 3 5t 2 dt . (6) Find d x4 10 2 sin t dt . dx (7) Find d x cost e dt . dx 3 Check your answer on your TI-89 by entering: d( (e^(cos(t)), t, 3, (x)), x) (8) Find d dx g(x) a f (t) dt . Check your answer on your calculator. IMSA Euler5.2 S 12