The graph of a(x) consist y = 80x Part 1 of 3 on the interval [4, 12], the graph is a Beis r= By 4 . (Gave the mum Part 2 of 3 A circle of radius / has an area A - 2172 [mri and so our sen harefore. "(x) x - Sufernil SNje (Kau cannot come back) Tutorial Exercise * (x) dx = 9 and fx) dx = 4.8, find ( "(x) dx. Part 1 of 3 We know that for a s bsc. [" rx) ax + (" mox) ax = ['n(x) ex. Since we have 2 s 4 5 6. ("x) ax + ( x x-2 Part 2 of 3 Therefore, "nx) ex - LRxox- 2 Part 3 of 3 Now , we can say the following . Tutorial Exercise Evaluate the integral. Part 1 of 4 An anti-derivative of kx", as long as ne -1. is Part 2 of 4 iative of nu) = 2 + 3 06 - 2 09 is since }(#) -and 1( 1. ) -0 . we now have To evaluate, we substitute s and 0 into the rivative F(u] - 2 4+ 75 45 - wl0 and subtract the results F(1) - F(0). we calculate F(1) and F(0) as follows. (1) - 2 + 25- Tutorial Exercise Evaluate the integral. 25 x dx Step 1 To find an antiderivative of vy, we will first rewrite the fraction as Vy - = x 2 1/2 _ 2x2 -1/2 Step 2 13/2 3/2 1/2 Simplifying gives Submit Skip (you cannot come back) Tutorial Exercise Evaluate the integral. 4 sec e tan e d6 Step 1 To find an antiderivative for Re) = 4 sec 8 tan 8, remember that sec o tan 8 = sec 8 sex(@) Therefore, an antiderivative for (8) = 4 sec 8 tan 8 is F(0) - 4 sec 8 4 sec(@) Step 2 We now have the following . 4 sec 0 tan o do - [4 sec 0]mis = 4 sec # - 4 sec 0 Submit Skip (you cannot come back) Tutorial Exercise Evaluate the integral . (3 + 2y)2 dy Part 1 of 5 To find an antiderivative of (3 + 2y)?, we will first expand the expression to obtain (3 + zy)? - 9 9 + 12 4 _ 12 y + a v 4 x Part 2 of 5 An antiderivative of kx", as long as n * -1, is k.* xntly Anti nti anti Part 3 of 5 Therefore, an antiderivative of fly) = 9 + 12y + 4y? is F(v) - 9y + X2 + 4 1 Submit Skip (you cannot come back)