HIGH SCHOOL ROUND ONE ⇒ You will have two minutes to evaluate each of the fifteen definite integrals that will be displayed one at a time on this screen. All answers must be simplified. At the end of the two minutes, all hands must go up and judges will grade your answers immediately. For each correct answer, you will receive one raffle ticket to be entered for prizes that will be drawn after dinner. At most five participants will move to the finals – to be determined by the total number of correct answers and tiebreaking criteria if necessary. Everyone moving to the Finals will receive $25. 2 0 1 5 U of S I NT E G RAT I ON B E E INTEGRAL #1 READY, GET SET,… 2:00 2 0 1 5 U of S I NT E G RAT I ON B E E INTEGRAL #1 ∫ π/6 ( ) sin(2x) + cos(6x) dx 0 2:00 2 0 1 5 U of S I NT E G RAT I ON B E E INTEGRAL #1 ∫ π/6 ( ) sin(2x) + cos(6x) dx 0 [ ]π/6 cos(2x) sin(6x) = − + 2 6 0 = 1 4 2 0 1 5 U of S I NT E G RAT I ON B E E INTEGRAL #2 READY, GET SET,… 2:00 2 0 1 5 U of S I NT E G RAT I ON B E E INTEGRAL #2 ∫1 (x − 2)(x − 1)(x + 1)(x + 2) dx 0 2:00 2 0 1 5 U of S I NT E G RAT I ON B E E INTEGRAL #2 ∫1 (x − 2)(x − 1)(x + 1)(x + 2) dx 0 ∫1 (x2 − 4)(x2 − 1) dx = 0 ∫1 (x4 − 5x2 + 4) dx = 0 ]1 5x 38 x − + 4x = = 5 3 15 0 [ 2 0 1 5 3 5 U of S I NT E G RAT I ON B E E INTEGRAL #3 READY, GET SET,… 2:00 2 0 1 5 U of S I NT E G RAT I ON B E E INTEGRAL #3 ∫1 xe2x dx 0 2:00 2 0 1 5 U of S I NT E G RAT I ON B E E INTEGRAL #3 ∫1 xe2x dx 0 [ integrate by parts: [ = ] 2x 1 xe 2 [ − 0 1 2 e xe = − 2 4 2 0 1 5 U of du = dx ∫1 , dv = e dx ] v = 21 e2x e2x dx 0 ] 2x 1 2x u=x 2x 0 S e2 + 1 = 4 I NT E G RAT I ON B E E INTEGRAL #4 READY, GET SET,… 2:00 2 0 1 5 U of S I NT E G RAT I ON B E E INTEGRAL #4 ∫ 1 (√ ) √ 4 5 5 x + x4 dx 0 2:00 2 0 1 5 U of S I NT E G RAT I ON B E E INTEGRAL #4 ∫ 1 (√ ) √ 4 5 5 x + x4 dx 0 ∫1 ( = 5/4 x 4/5 +x ) dx 0 [ = ] 9/5 1 9/4 4x 9 + 5x 9 0 = 1 2 0 1 5 U of S I NT E G RAT I ON B E E INTEGRAL #5 READY, GET SET,… 2:00 2 0 1 5 U of S I NT E G RAT I ON B E E INTEGRAL #5 ∫ π/4 0 sin x dx (1 + sin x)(1 − sin x) 2:00 2 0 1 5 U of S I NT E G RAT I ON B E E INTEGRAL #5 ∫ π/4 0 sin x dx (1 + sin x)(1 − sin x) ∫ π/4 = 0 sin x dx = 2 1 − sin x ∫ √2/2 =− 1 ∫ π/4 0 sin x dx 2 cos x [ 1 du u = cos x, u2 ] du = − sin x dx [ ]√2/2 √ 1 = = 2−1 u 1 2 0 1 5 U of S I NT E G RAT I ON B E E INTEGRAL #6 READY, GET SET,… 2:00 2 0 1 5 U of S I NT E G RAT I ON B E E INTEGRAL #6 ∫2 √ 3 −3 x+3 dx 5 2:00 2 0 1 5 U of S I NT E G RAT I ON B E E INTEGRAL #6 ∫2 √ 3 −3 x+3 dx 5 ∫1 =5 0 [ [ x+3 1/3 , u dx u = 5 ] 4/3 1 15u = 4 = ] 1 du = dx 5 0 15 4 2 0 1 5 U of S I NT E G RAT I ON B E E INTEGRAL #7 READY, GET SET,… 2:00 2 0 1 5 U of S I NT E G RAT I ON B E E INTEGRAL #7 ∫ 4/π 3/π 1 1 1 sec tan dx 2 x x x 2:00 2 0 1 5 U of S I NT E G RAT I ON B E E INTEGRAL #7 ∫ 4/π 3/π 1 1 1 sec tan dx 2 x x x ∫ π/4 =− π/3 [ 1 sec u tan u du u = , x ] 1 du = − 2 dx x [ ]π/4 = − sec u π/3 = 2− 2 0 1 5 √ 2 U of S I NT E G RAT I ON B E E INTEGRAL #8 READY, GET SET,… 2:00 2 0 1 5 U of S I NT E G RAT I ON B E E INTEGRAL #8 ∫ ln 3 0 ex dx x 3 (1 + e ) 2:00 2 0 1 5 U of S I NT E G RAT I ON B E E INTEGRAL #8 ∫ ln 3 0 ex dx x 3 (1 + e ) ∫4 = [ u−3 du u = 1 + ex, du = ex dx ] 2 [ ]4 1 = − 2 2u 2 3 = 32 2 0 1 5 U of S I NT E G RAT I ON B E E INTEGRAL #9 READY, GET SET,… 2:00 2 0 1 5 U of S I NT E G RAT I ON B E E INTEGRAL #9 ∫ π/4 (tan3 x + 1)(tan2 x + 1) dx 0 2:00 2 0 1 5 U of S I NT E G RAT I ON B E E INTEGRAL #9 ∫ π/4 (tan3 x + 1)(tan2 x + 1) dx 0 ∫ π/4 (tan3 x + 1) sec2 x dx = 0 ∫1 = [ (u3 + 1) du u = tan x, ] du = sec2 x dx 0 [ 4 u = +u 4 2 0 1 5 U ]1 0 of 5 = 4 S I NT E G RAT I ON B E E INTEGRAL #10 READY, GET SET,… 2:00 2 0 1 5 U of S I NT E G RAT I ON B E E INTEGRAL #10 ∫1 √ 2/2 ( )2 1 1 1 + 2 dx 3 x x 2:00 2 0 1 5 U of S I NT E G RAT I ON B E E INTEGRAL #10 ∫1 √ 2/2 ( )2 1 1 1 + 2 dx 3 x x [ 1 1 =− u2 dx u = 1 + 2 , 2 3 x [ 3 ]2 u = − 6 3 ∫2 ] 2 du = − 3 dx x 19 = 6 2 0 1 5 U of S I NT E G RAT I ON B E E INTEGRAL #11 READY, GET SET,… 2:00 2 0 1 5 U of S I NT E G RAT I ON B E E INTEGRAL #11 ∫ π2 √ sin x √ dx π2/4 x 2:00 2 0 1 5 U of S I NT E G RAT I ON B E E INTEGRAL #11 ∫ π2 √ sin x √ dx π2/4 x [ ∫π sin u du =2 π/2 [ = − 2 cos u u= √ x, ] 1 du = √ dx 2 x ]π π/2 = 2 2 0 1 5 U of S I NT E G RAT I ON B E E INTEGRAL #12 READY, GET SET,… 2:00 2 0 1 5 U of S I NT E G RAT I ON B E E INTEGRAL #12 ∫ √2 √ x4 − x2 dx 1 2:00 2 0 1 5 U of S I NT E G RAT I ON B E E INTEGRAL #12 ∫ √2 √ x4 − x2 dx 1 ∫ √2 √ √ ∫ √2 √ = x2 x2 − 1 dx = x x2 − 1 dx 1 1 = 2 [ = 1 ∫1 √ [ u du 2 u = x − 1, ] du = 2x dx 0 ] 3/2 1 u 3 2 0 1 5 = 0 U of 1 3 S I NT E G RAT I ON B E E INTEGRAL #13 READY, GET SET,… 2:00 2 0 1 5 U of S I NT E G RAT I ON B E E INTEGRAL #13 ∫ 13 6 1 dx 5/3 (x − 5) 2:00 2 0 1 5 U of S I NT E G RAT I ON B E E INTEGRAL #13 ∫ 13 6 1 dx 5/3 (x − 5) ∫8 u−5/3 = [ u = x − 5, ] du = dx 1 ]8 [ 3 = − 2/3 2u 1 9 = 8 2 0 1 5 U of S I NT E G RAT I ON B E E INTEGRAL #14 READY, GET SET,… 2:00 2 0 1 5 U of S I NT E G RAT I ON B E E INTEGRAL #14 ∫3 0 x √ dx x+1 2:00 2 0 1 5 U of S I NT E G RAT I ON B E E INTEGRAL #14 ∫3 0 x √ dx x+1 [ u= ∫2 = 1 √ x + 1, ] 2 (u2 − 1) · 2u du u [ ∫2 u3 2 −u = 2 (u − 1) du = 2 3 1 2 0 1 5 2u du = dx u = x + 1, U of S ]2 1 8 = 3 I NT E G RAT I ON B E E INTEGRAL #15 READY, GET SET,… 2:00 2 0 1 5 U of S I NT E G RAT I ON B E E INTEGRAL #15 ∫3 1 x2016 − x2014 dx 2015 2014 x +x 2:00 2 0 1 5 U of S I NT E G RAT I ON B E E INTEGRAL #15 ∫3 1 x2016 − x2014 dx 2015 2014 x +x ∫3 = 1 x2014(x2 − 1) dx x2014(x + 1) ∫3 x2014(x + 1)(x − 1) = dx 2014 x (x + 1) 1 [ 2 ]3 ∫3 x = (x − 1) dx = −x = 2 2 1 1 2 0 1 5 U of S I NT E G RAT I ON B E E THANKS FOR PLAYING LET’S EAT! (YOU HAVE TWO MINUTES TO FINISH YOUR FOOD) THE FINAL ROUND BEGINS AFTER DINNER 2 0 1 5 U of S I NT E G RAT I ON B E E
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