MATH 1207 R02 MIDTERM EXAM 2 SOLUTION
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1 MATH 7 R MIDTERM EXAM SOLUTION FALL 6 - MOON Name: Write your answer neatly and show steps. Except calculators, any electronic devices including laptops and cell phones are not allowed. () (5 pts) Find the derivative of g(x) = ln(x + ). Let y = g(x) and u = x +. Then y = ln u. Setting u = x + : pts. dy dx = dy du du dx = u x = x x +. By applying the chain rule, getting the answer x x + : 5 pts. Date: October 8, 6.
2 MATH 7 R Midterm Exam () (a) (6 pts) Evaluate the limit e x sin x lim x 4x e x sin x lim. x 4x x e x cos x 8x x e x + sin x 8 = 8 Fall 6 - Moon e x cos x Applying l Hospital s rule correctly and obtaining lim : 4 x 8x pts. Applying l Hospital s rule once again and getting the answer 8 : 6 pts. (b) (6 pts) Evaluate the limit ( ) ln lim ( + t) 3 t ln t t Therefore lim ( + t) 3 t. t ( ) ( + t) 3 t t 3 ln( + t) t 3 t t t lim ( + t) 3 t = e 6. t 3 +t ln( + t) 6 t + t = 6 Knowing the technique to take the logarithm of the limit: pts. Using l Hospital s rule and obtaining the limit 6: 5 pts. Getting the answer e 6 : 6 pts.
3 MATH 7 R Midterm Exam Fall 6 - Moon (3) (8 pts) A fossilized animal bone is unearthed on the ruins of Pompeii. It contains 8% of Carbon-4 found in living matter. About how old is the bone? (It takes 573 years to be half of the initial amount of Carbon-4.) Let m(t) be the amount of Carbon-4 in the bone after t years. Then m(t) = Ce kt Ce 573k = m(573) = C e573k = 573k = ln k = ln 573 m(t) = Ce ln 573 t m(t) =.8C.8C = Ce ln 573 t.8 = e ln 573 t ln 573 ln.8 t = ln.8 t = 573 ln years Writing correct general solution m(t) = Ce kt for the amount of Carbon-4: pts. Finding m(t) = Ce ln 573 t : 5 pts. Writing the answer with an appropriate unit years: 8 pts. Writing the answer without stating the unit: - pt. 3
4 MATH 7 R Midterm Exam Fall 6 - Moon (4) (a) (6 pts) Evaluate the definite integral e x 4 ln xdx. x 4 ln xdx = ln x 5 x5 e u = ln x, dv = x 4 dx du = x dx, v = 5 x5 5 x5 x dx = 5 x5 ln x x 4 ln xdx = 5 x5 ln x 5 x5 ] e = ( 5 e5 ln e 5 e5 5 x4 dx = 5 x5 ln x 5 x5 + C ) ( 5 ln ) 5 = 4 5 e5 + 5 Finding appropriate u = ln x and dv = x 4 dx: pts. By using integration by parts, getting the antiderivative 5 x5 ln x 5 x5 + C: 5 pts. Getting the answer 4 5 e5 + 5 : 6 pts. (b) (6 pts) Evaluate the indefinite integral 6x dx. 4x 4 Set u = x. Then du = 4xdx. 3 6x dx = 6x 4x 4 (x ) dx = du u = 3 sin u + C = 3 sin (x ) + C Finding an appropriate substitution u = x : pts. Getting the answer 3 sin (x ) + C: 6 pts. 4
5 MATH 7 R Midterm Exam Fall 6 - Moon (5) For x >, let f(x) = tan (x) + tan (/x). (a) (5 pts) Find f (x) and simplify the answer as much as you can. f (x) = + x + + ( ) ( x ) = + x x + =. Knowing the derivative of tan x: pts. Getting the answer f (x) = : 5 pts. x (b) (4 pts) By using (a), compute f( ) without using calculator. (Unless you justify your answer, you don t get any credit.) Because f (x) =, f(x) is a constant function. Therefore f( ) = f() = tan () + tan () = π 4 + π 4 = π. Mentioning that f(x) is a constant function: pts. Getting f( ) = π : 4 pts. Unless use the fact that f (x) = and f(x) is a constant, one cannot get any credit. (c) (4 pts) Explain the conclusion of (a) and (b) without using Calculus. (Hint: Draw a right triangle.) For any right triangle whose two sides are and x, A = tan (x) and B = tan (/x). Because A + B = π, we obtain the result. Knowing the definition of tan x: pt. Providing a correct explanation: 4 pts. Explaining (b) only: 3 pts. 5
6 MATH 7 R Midterm Exam Fall 6 - Moon (6) ( pts) By using the midpoint rule with n = 5, find an approximation of the integral to four decimal places. sin(x )dx x = 5 =. x =, x =., x =.4, x 3 =.6, x 4 =.8, x 5 = x = x + x =., x =.3, x 3 =.5, x 4 =.7, x 5 =.9 sin(x )dx = sin( x ) x + sin( x ) x + sin( x 3) x + sin( x 4) x + sin( x 5) x = sin(. ) x + sin(.3 ) x + sin(.5 ) x + sin(.7 ) x + sin(.9 ) x ( (.45466)..58 Finding x =. and the coordinates of x i and x i : 4 pts. Writing the formula sin(. ) x + sin(.3 ) x + sin(.5 ) x + sin(.7 ) x + sin(.9 ) x for the approximation: 8 pts. Getting the answer.58: pts. 6
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