Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function. G (x) = cos vzt dt Part 1 of 3 By the Fundamental Theorem of Calculus Part 1, we know that if 9(x) = "f(t) at, then g'(x) = F(x). O P g'(x) = f(x). g' (x ) = 0. g' ( x ) = f ' ( x ) . g' ( x ) = a. Part 2 of 3 We know that if A(x ) dx - A , then (x ) dx - ~ - Part 3 of 3 Therefore, cost 2t at = - "cost 2t dt. Let f(t) = cosv2t, g(x) = cosvzt at, and G(x) - 'cosvit at. Now, by FTC1, we know that G'(x) = -g'(x) = -f(x) = Tutorial Exercise Evaluate the integral. 16 ( 2 + 3 u# - 248) du Part 1 of 4 n+1 An+ 1 An anti-derivative of kx", as long as n # -1, is kx n+ 1 Pn+1 Part 2 of 4 Therefore, an antiderivative of f(u) = 2 + = u4 - 3 us is F(u) = 2 Submit Skip (you cannot come back) Tutorial Exercise Evaluate the integral. [25 x - 2 dx VX Step 1 To find an antiderivative of -2 Vy we will first rewrite the fraction as -= x 1/2 | 1/2 -2x -1/2 -1/2 Step 2 Now , 125 2 dx = x3/2 - 2x-3/2 ax - - Submit Skip (you cannot come back ) Tutorial Exercise Evaluate the integral. 4 sec 0 tan 0 de Step 1 To find an antiderivative for f(0) = 4 sec 0 tan 0, remember that sec 0 tan 0 = - Therefore, an antiderivative for f(0) = 4 sec 0 tan 0 is F(0) = Submit Skip_(you cannot come back). Tutorial Exercise Evaluate the integral. 1 3 (3 + 24 ) 2 dy Part 1 of 5 To find an antiderivative of (3 + 2y)2, we will first expand the expression to obtain (3 + 2y)2 = 9 9 + 12 0 _ 12 y+4 4 x2. Part 2 of 5 An antiderivative of kx", as long as n # -1, is k Submit Skip (you cannot come back)