Tutorial Exercise Evaluate the integral . 6 ( 2 + But - 208) du Part 1 of 4 An anti-derivative of kx", as long as n # -1, is kx n+ 1 0 n+1 Part 2 of 4 Therefore, an antiderivative of A(u) = 2 + = u4 - 2 us is F ( U ) = 2 \\u V u + 2 4 2 5 3 410 10 5 2 5 4 10 10 Part 3 of 4 since 3 (5 ) = 2/25 2 2/25 and * ( 70 ) - 3/40 3/40 , we now have ( 2 + 3 14 - 2 09 = [20+ 25 45 - 2 41072 To evaluate, we substitute 1 and 0 into the antiderivative F(u) = 2 u + 5 us - - ulo and subtract the results F(1) - F(0). We calculate F(1) and F(0) as follows. F(1) = 2 + 25 - 3/40 3/40 = 401/200 401/200 F(0) = 0 0 Part 4 of 4 Combining these results, we have the following . = F(1) - F(0) Submit Skip (you cannot come back) Evaluate the integral. 25x -2 dx Step 1 To find an antiderivative of 2 Vx , we will first rewrite the fraction as -= x 1/2 1/2 _ 2x 1/2 -1/2 Step 2 (25 x - 2 dx B/2 2 3/2 2 1/2 1 1/2 7125 13/2 2 3/2 1/2 Step 3 Simplifying gives 2/3 - 2/3 x 3/2 - 4 4 x 1 / 2 / 25 Step 4 Remember that 251/2 = v25 = 5 and that abic = (al/c)b in order to evaluate the following. Vidx = [3x3/2 - 4x1/2125 - (3 - ) -(3-4) Tutorial Exercise Evaluate the integral. (3 + 24 ) 2 dy Part 1 of 5 To find an antiderivative of (3 + 2y), we will first expand the expression to obtain (3 + 2y)2 = 9 9 + 12 0 _ 12 +4 4x. Part 2 of 5 nti on+ 1 An antiderivative of kx", as long as n # -1, is kX Part 3 of Therefore, an antiderivative of f(y) = 9 + 12y + 4y2 is F(Y ) = 9y + 6 6 12 + 4 y x 3 Part 4 of 6 We now have (9 + 12 + 472) dy = [ox + 6x2 + 3x ]]. To evaluate, we substitute 4 and 3 into the antiderivative F(y) = 9y + 6y2 + 43 and subtract the results F(4) - F(3). F ( 4 ) = 36 + 96 + F(3) = Submit Skip_(you cannot come back)