Questions:

3 s 2. (3 points each] Suppose f is an odd function with f f(3:) d3: = T and / x) d3: = 3. o 0 Evaluate each of the following denite integrals. (a) rm are (b) jinx) as: (c) fire) dz: 4. (5 points each) For each function below, determine if the function satises the hypothesis of the Mean Value Theorem for Integrals. If so1 determine the value(s) of a guaranteed by the conclusion of the Mean Value Theorem for Integrals. (a) y = 1 |:r| on [1,]] (b) y = :52 4 on [3,0] Example. Compute each of the following derivatives. 1. a Vi + 3 dt = Jx 2 + 3 the lower bound does not matter ! ( as long as it is a constant ) 2. dx d viz + 3 dt = JX 2 + 3 - going backwards - negate + switch bounds first 3. d X + 3 dx J "vetsat = a ( - s* Jt z + 3 dt ) = - g ( h(x) ) where g (x 1 = Jy J + 2 + 3 dt use Chain d "sin x h (x 1 = sinx nile 4. dx V12 + 3 dt dx ( g (n(xi) ) = g'( hixl) . h ' ( x ) = J[ sinx12 + 3 . cosx d fin x 5. dx J tan x da = tan ( Inx ) .* d In(x2+5x) 6. dx tan x dx = tan ( In ( x 2 + 5x) ) ( x2 + 5 X (2x +5 ) 2Theorem (Fundamental Theorem of Calculus, Part 2). If f is continuous on [a, b], then [ s( ) da = F(x) |a = Flb)-F(a) Where F(x) is an and derivative of fix). L can be any antiderivative - don't need " + c Example. Compute each of the following integrals. - 5 1. L ( 307 - 4 ) as = x3 - 4 1/x1 + C " + ( " = (1 - 5 ) 3 - 4in1- 51 + ( ) - (1 -113 - 4 /-11+ ( ) will always In( 1) = 0 cancel = -125- 41n(5) +1 =-124-41n(5) we don't need to include it 2 . yz - y+1 dy 4 2 - 4 + 1 Vy dy = [ 4 / y 312 - y12 + y/3) dy - 15 12 512 1 12 4 + $12 - 2 y' 12 + 2 4 12 4 2 2 (54 ) +254 ) - ( 1 ) 512 W / N ( 1 ) 312 + 2 ( 1) 12 = F (32 ) - 2 (8 ) + 4 - 5 3 3 2+ 2) = 42 _4+2 = 184- 70 + 30 = 146 5 3 15 IS 15 15Integrals of Even and Odd Functions. If f(x) is even, then ... Safexax = Sa fixlax , so Ja fixlax = 2 So fix) dx Fact : If f(x) is odd, then ... So fixl dx = - Ja fixidx , so Ja fixlax = 0. sin ( - x ) = - sinx Cos ( - X ) = COSx Example. Use symmetry to compute (23 + tan(Tz*) - 12x35 ) da. fan (- x ) = - tany = J -5 x 2 dx + _ tan( I(x 3 ) dx - 12 x 35 dx 0 even tan ( +1 ( - * ) ? ) = tan ( - mix ? ) 12 ( - x ) 135 = - 12x35 = - tan ( 7(x 3 ) + odd odd = 2 8 x2 dx = 2x3 15 - 2 ( 125 ) - 0= 250 3 3 3 Average Value of a Function. The average of a list of numbers q1, 92, . .., In is given by... Fit qz t ... + En = n For a continuous function f(x) on an interval [a, b], we could estimate the average value of f(x) by sampling f (r) at many r-values that are evenly spaced apart. h x-values that are AX apart : y = f(x) XIJ X 2 , ... , Xn 4X = 6-9 n Average ~ fix , ) + fix z ) + .. + f(Xn ) h = [, fixi) 4x = 6-a AX El fix : )Ax = 1= 1 n n 4X b -a Ax in = b - aQuestion. Does a function always achieve its average value on an interval? No! Must be continuous EX : If fix ) f = To So fixlav = 2 1/2 1 / 2 But fixit ? Theorem (Mean Value Theorem for Integrals). If f(x) is a continuous function on the interval [a, b], then ... there is an X-valve c in (aib) so that f (c ) = f = b-a Ja fixI dx (Recall MIVT : avg. v. o. c. = instantaneous r.o.c. ) Extra Example. Find all r-values c guaranteed by the Mean Value Theorem for Integrals for the function f(x) = 1 - - 2 over the interval [0, a] where a is a positive real number. f(x ) = 1- x2 is continuous everywhere - cts. on To, a] - MUT for Integrals applies ! f = a-o Jo ( 1 - x 2 ) dx = = (x - XS 3 a2 - -la - as 272 ) - a (0 ) = 2 ( a - 9 ) = 2 3 = W / N f(x) = f - x 2 = I a 2 3 X = + =+ 3 1- X2 = WIN a 2 40 2 X 2 C = 3 positive only