Math 8, Exam, Study Guide Problem Solution. Evaluate xe x dx. Solution: We evaluate the integral using the u-substitution method. Let u x. Then du xdx du xdx and we get: xe x dx e x xdx e u du e u du eu + C ex + C
Math 8, Exam, Study Guide Problem Solution. Evaluate 6x + dx. Solution: We evaluate the integral using the u-substitution method. Let u 4x. Then du 4dx du dx and we get: 4 6x + dx (4x) + dx u + 4 du 4 u + du 4 arctan u + C 4 arctan(4x)+c
Math 8, Exam, Study Guide Problem 3 Solution 3. Evaluate t 3 t4 +9 dt. Solution: Weevaluatetheintegralusingtheu-substitution method. Let u t 4 +9. Then du 4t 3 dt 4 du t3 dt and we get: t 3 t4 +9 dt 4 t4 +9 t3 dt u 4 du du 4 u [ ] u + C u + C t4 +9+C
Math 8, Exam, Study Guide Problem 4 Solution 4. Evaluate tan θ sec θdθ. Solution: Weevaluatetheintegralusingtheu-substitution method. Let u tanθ. Then du sec θdθ and we get: tan θ sec θdθ u du 3 u3 + C 3 tan3 θ + C
Math 8, Exam, Study Guide Problem 5 Solution 5. Evaluate x x dx. Solution: Weevaluatetheintegralusingtheu-substitution method. Let u x.then du xdx du xdx. The limits of integration becomes u and u. Weget: x x dx x (xdx) ( u ) du u / du u / du [ ] 3 u3/ [ ] 3 ()3/ [ ] 3 ()3/ 3
Math 8, Exam, Study Guide Problem 6 Solution 6. Evaluate π π cos x 4+3sinx dx. Solution: We evaluate the integral using the u-substitution method. Let u 4+3sinx. Then du 3cosxdx du cosxdx. The limits of integration become u 4+ 3 3sin( π) 4andu 4+3sinπ 4. Weget: π π cos x 4+3sinx dx π π 4 4 because the limits of integration are identical. 4+3sinx cos xdx u 3 du
Math 8, Exam, Study Guide Problem 7 Solution 7. Evaluate 4 x dx. Solution: Theintegrandisundefinedatx. Therefore,thisisanimproperintegral. We evaluate the integral by turning it into a limit calculation. 4 dx lim x b b [ b lim 4 x dx 4arcsinx ] b lim [4 arcsin b 4arcsin] b 4arcsin 4arcsin ( π ) 4 4() π We evaluated the limit lim 4arcsinb by substituting b using the fact that f(b) b 4arcsinb is left-continuous at b.
8. Find dy for each of the following: dx x (a) y +t dt (b) y (c) y x tan x sin ( t ) dt Math 8, Exam, Study Guide Problem 8 Solution dt (hint: when you simplify, you will get a constant) +t Solution: Inallparts,weusetheFundamentalTheoremofCalculusPartII: (a) The derivative is: d dx g(x) a f(t) dt f(g(x))g (x) dy dx d dx x +t dt +(x) (x) +x (b) The derivative is: dy dx d dx sin x ( ( x ) ) sin ( t ) dt ( x ) sinx x (c) The derivative is: dy dx d tan x dx +t dt (tan x) +(tanx) +tan x sec x sec x sec x
Math 8, Exam, Study Guide Problem 9 Solution 9. Find the area of the region enclosed by the curves y x and y x +4x. Solution: y 4 3 x The formula we will use to compute the area of the region is: Area b a (top bottom) dx where the limits of integration are the x-coordinates of the points of intersection of the two curves. These are found by setting the y s equal to each other and solving for x. y y x 4x x(x ) x x +4x x,x From the graph we see that the top curve is y x +4x and the bottom curve is y x.
Therefore, the area is: Area b a (top bottom) dx [( x +4x ) ( x )] dx ( 4x x ) dx [ x 3 x3 ] () 3 ()3 8 3
Math 8, Exam, Study Guide Problem Solution. Find the area of the region enclosed by the curves x + y andx +3y. Solution: y - x - The formula we will use to compute the area of the region is: Area d c (right left) dx where the limits of integration are the y-coordinates of the points of intersection of the two curves. These are found by setting the x s equal to each other and solving for y. x x y 3y + y (y ) (y +)(y ) y, y From the graph we see that the right curve is x 3y +andthe leftcurve isx y.
Therefore, the area is: Area d c (right left) dx [( 3y + ) ( y )] dy ( y ) dy [ y ] 3 y3 [() 3 ] ()3 [( ) 3 ] ( )3 8 3
Math 8, Exam, Study Guide Problem Solution. Find the volume of the following solid: The base of the solid is the circle x + y. The cross-sections are isosceles right triangles perpendicular to the y-axis. Solution: The volume formula we will use is: V Z d c A(y) dy where A(y) is the cross sectional area of the solid as a function of y. We choose to integrate with respect to y in this problem because the cross sections are perpendicular to the y-axis. To determine the function A(y) we first note that the cross sections are isosceles right triangles. The triangle may be oriented in one of two ways: () with the hypotenuse lying inside the circle x + y or () with one of the sides lying inside the circle. Let s assume () for the moment. In this case, the length of the hypotenuse is x where x p y after solving the equation x + y forx. The other two sides of the triangle are equal since the triangle is isosceles. Letting the other sides be a, we use the Pythagorean Theorem to find a. a + a (x) a 4x a 4( y ) a ( y ) a p ( y )
The area of the triangle is: A(y) bh (a)(a) a y y The cross sections start at c and end at d. Therefore, the volume is: V Z apple y apple y dy 3 y3 3 ()3 3 + 3 apple ( ) 3 ( )3 4 3 Now let s assume (), that one of the sides lies in the circle. In this case, the length of the side is x where x p y. The other side of the triangle is also x since the triangle is
isosceles. The area of the triangle is then: A(y) bh (x)(x) x ( y ) The cross sections start at c and end at d. Therefore, the volume is: V Z apple y apple y dy 3 y3 apple 3 ()3 ( ) 3 + 3 3 ( )3 8 3 3
Math 8, Exam, Study Guide Problem Solution. Find the volume of the solid generated by revolving the region bounded by y x, y,andx aboutthe: (a) x-axis (b) y-axis Solution: (a) We find the volume of the solid generated by revolving around the x-axis using the Washer Method. Thevariableofintegrationisx and the corresponding formula is: V π b a [ (top) (bottom) ] dx The top curve is y andthebottomcurveisy x.thelowerlimitofintegration is x. Theupperlimitisthex-coordinate of the point of intersection of the curves y andy x.tofindthis,wesetthey s equal to each other and solve for x. y y x x 4 The volume is then: V π 4 4 [ () ( x ) ] dx π (4 x) dx [ π 4x ] 4 x π [4(4) ] (4) 8π
y 3 4 x (b) We find the volume of the solid generated by revolving around the y-axis using the Shell Method. Thevariableofintegrationisx and the corresponding formula is: V π b a x (top bottom) dx The top and bottom curves are the same as those in part (a). So are the limits of integration. The volume is then: V π 4 4 x ( x ) dx ( π ) x x 3/ dx [ π x ] 4 5 x5/ π [4 5 ] (4)5/ 3π 5
Math 8, Exam, Study Guide Problem 3 Solution 3. Find the volume of the solid generated by revolving the region bounded by the y-axis and the curve x,for y 4, about the: y (a) x-axis, using the Method of Cylindrical Shells (b) y-axis Solution: (a) We find the volume of the solid obtained by rotating about the x-axis using the Shell Method. Thevariableofintegrationisy and the corresponding formula is: V π d c y (right left) dy The right curve is x y and the left curve is x (they-axis). The volume is then: V π π 4 4 ( ) y y dy dy [ ] 4 π y π [(4) ()] π y 4 3 x
(b) We find the volume of the solid obtained by rotating about the y-axis using the Disk Method. Thevariableofintegrationisy and the corresponding formula is: V π where f(y) y.thevolumeisthen: d c f(y) dy 4 V π π 4 ( ) dy y 4y dy [ 4π ] 4 y [( 4π ) 4 3π ( )]
Math 8, Exam, Study Guide Problem 4 Solution 4. In some chemical reactions, the rate at which the amount of asubstancechangeswith time is proportional to the amount present. Consider a substance whose amount obeys the equation: dy dt.6y where t is measured in hours. If there are grams of the substance present when t, how many grams will be left after hour? Solution: Theamountofthesubstancey(t) isgivenbytheformula: y(t) y e.6t where y gramsistheinitialamountofthesubstance. Afterhour, the amount of the substance is: y() e.6 grams
Math 8, Exam, Study Guide Problem 5 Solution 5. Suppose the rate at which the number of people infected with a disease dy is proportional dt to the number of people currently infected y: dy dt ky Suppose that, in the course of any given year, the number of people infected is reduced by %. If there are, infected people today, how many years willittaketoreducethe number to? Solution: Thenumberofpeopleinfectedisgivenbythefunction: y(t) y e kt where y, is the initial number of people infected. To answer the question in the problem, we need to find the value of k. Sincethenumberofpeopleinfectedisreducedby % in the course of any given year, the number of people infected after the first year is:,.(, ) 8, This corresponds to the value y(). Using the function above for y(t) we get: y(), e k() 8,, e k e k e k 4 5 8,, k ln 4 5 To find how many years it will take for the number of infected people to reduce to,, we set y(t) equal to, and solve for t. y(t), e (ln 4 5 )t,, e (ln 4 5 )t,, 4 e(ln 5 )t 4 e(ln 5 )t ln ( ln 4 ) t 5 t ln ln 4 5
Math 8, Exam, Study Guide Problem 6 Solution 6. Consider the definite integral: 4 ( x + x ) dx (a) Compute the exact value of the integral. (b) Estimate the value of the integral using the Trapezoidal Rule with N 4. (c) Estimate the value of the integral using Simpson s Rule with N 4. (d) Which of the above two estimate is more accurate? Solution: (a) The exact value is: 4 ( x + x ) dx [ 3 x3 + ] 4 x 3 (4)3 + (4) 88 3 (b) Using N 4,thelengthofeachsubintervalof[, 4] is: The estimate T 4 is: x b a N 4 4 T 4 x [f() + f() + f() + f(3) + f(4)] [ ( +)+( +)+( +)+(3 +3)+(4 +4) ] [ + 4 + + 4 + ] 3
(c) The estimate S 4 is: S 4 x [f() + 4f() + f() + 4f(3) + f(4)] 3 [ ( +)+4( +)+( +)+4(3 +3)+(4 +4) ] 3 [ + 8 + + 48 + ] 3 88 3 (d) Clearly, S 4 is more accurate because it is the exact value of the integral.
Math 8, Exam, Study Guide Problem 7 Solution 7. Consider the definite integral: Estimate the value of the integral using: (a) the Trapezoidal Rule with N (b) the Midpoint method with N (c) Simpson s Rule with N 4 dx +x Solution: (a) The length of each subinterval of [, ] is: The estimate T is: x b a N T x [f() + f() + f()] [ + + + + ] + [ ++ ] 5 (b) The length of each subinterval of [, ] is x justasinpart(a). TheestimateM is: [ ( ) ( )] 3 M x f + f [ 4 5 + 4 3 7 65 + ( ) + + ( 3 ) ]
(c) We can use the following formula to find S 4 : S 4 3 M + 3 T where M and T were found in parts (a) and (b). We get: S 4 ( ) 7 + ( ) 3 65 3 48 65 + 3 43 39