A. Incorrect! The letter t does not appear in the expression of the given integral

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1 AP Physics C - Problem Drill 1: The Funamental Theorem of Calculus Question No. 1 of 1 Instruction: (1) Rea the problem statement an answer choices carefully () Work the problems on paper as neee (3) Question 1. In the integral f( u) u, where a is a fie number, the variable of integration is a Question #1 (A) t (B) (C) a (D) u (E) f The letter t oes not appear in the epression of the given integral In the given integral, is rather the upper limit of integration. In the given integral, a is rather the lower limit of integration. Inee, the variable with respect to which integration is performe is u. In the given integral, f is the name of the function. Here the problem consists in looking for the variable with respect to which integration is performe. This variable is the one that appears uner the integration symbol. In this question this variable turns out to be u.

2 Question No. of 1 Instruction: (1) Rea the problem statement an answer choices carefully () Work the problems on paper as neee (3) Question # Question. The function t y = e t is. (A) Strictly ecreasing on (, ). (B) Strictly increasing on (, ). (C) Not ifferentiable. (D) Not continuous. (E) Constant Accoring to the Funamental Theorem of Calculus, y in this case remains strictly positive for any. Therefore, y is strictly increasing on the whole real line. B. Correct! Accoring to the Funamental Theorem of Calculus, y remains strictly positive for any. Therefore y is strictly increasing on the whole real line. By the Funamental Theorem of Calculus, the given function is ifferentiable since the integran is continuous. D. Incorrect! By the Funamental Theorem of Calculus, the given function is ifferentiable since the integran is continuous. A ifferentiable function is a fortiori continuous. Accoring to the Funamental Theorem of Calculus, y in this case remains strictly positive for any. Therefore, y is strictly increasing on the whole real line. Since the erivative is strictly positive, this cannot be a constant function. By the Funamental Theorem of Calculus, -t y' = e t e = Recall that the variations of a ifferentiable function is stuie by looking at the sign of its first erivative. In our case the first erivative remains positive for any number on the real line. Therefore our function is strictly increasing on the real line.

3 Question No. 3 of 1 Instruction: (1) Rea the problem statement an answer choices carefully () Work the problems on paper as neee (3) Question 3. Accoring to the Funamental Theorem of Calculus, there is a function F with a erivative. Write an eplicit formula for F. The Formula shoul not an integral. Question #3 (A) (B) t t (C) (D) (E) None of these The propose formula containe an integral. The erivative of square over with respect to is. The erivative of negative square over with respect to is. By irect ifferentiation using the prouct rule, we may confirm the answer. The same result may be erive by the Funamental Theorem of Calculus. See below for etails. One of the given answers is correct. Please try again. By the Funamental Theorem of Calculus, an eample of a function with erivative is F( ) = t t. In orer to evaluate the last integral, we have to consier two cases. t Case 1: If, then F( ) = tt = = =. Case : I f <, then It can be reaily verifie that t F( ) = tt = = + =. F( ) =. A secon approach consists calculating F (). By the chain rule, we have F '( ) = ( ). + = + = + = You may note that we have use the fact that ( ) = for. The last result can be establish by noticing that =. Differentiating both sies of the last equation with respect to an solving the resulting equation for ( ) yiels the esire result.

4 Question No. of 1 Instruction: (1) Rea the problem statement an answer choices carefully () Work the problems on paper as neee (3) Question. Let a an b be two fie numbers. Let f be a function continuous on the real line. Define two new functions by F( ) = f( t) t an G ( ) = ftt ( ).The following assertion is correct a b Question # (A) ( F( ) ) ( G( ) ) b (B) G ( ) F ( ) = ftt ( ) a a (C) ( F( ) ) = ( G( ) ) an G() = f( t) t + F( ) b b (D) ( F( ) ) = ( G( ) ) an G() = f( t) t + F( ) a (E) a = b By the Funamental Theorem of Calculus F ()=f() an G ()=f() as well. Thus F ()=G (). The limits of integration are reverse on the right han sie of the equation. You may use the splitting property of efinite integrals to confirm this assertion. C. Correct! By the Funamental Theorem of Calculus, F ()=f() an G () = f(). Hence F () = G (). Net, by using the splitting property of efinite integrals, the secon part of the statement can be confirme as well. D. Incorrect! The secon part of the assertion is incorrect. The limits of integration are reverse on the right han sie. It cannot be inferre that a = b from the given information. By the Funamental Theorem of Calculus, F () = f() an G () = f(). Hence F () = G (). Net, a a G( ) = f () t t = f () t t + f () t t = f () t t + F( ) correct.. Therefore the secon part of the assertion is also b b a b

5 Question No. 5 of 1 Instruction: (1) Rea the problem statement an answer choices carefully () Work the problems on paper as neee (3) Question 5. Let sin( t) F( ) = F( ) t. Then lim t is equal to Question #5 (A) / (B) (C) 1 (D) -1 (E) This not a vali limit. First use the Funamental Theorem of Calculus an then fin the limit. C. Correct! By using the Funamental Theorem of Calculus an then applying what you know about trigonometric limits, the answer is 1. D. Incorrect! First use the Funamental Theorem of Calculus an then fin the limit. First use the Funamental Theorem of Calculus an then fin the limit. The propose limit satisfies the hypotheses for the application of l Hopital rule. Inee both the numerator an the enominator are ifferentiable functions an they vanish as approaches. Therefore sin( ) F( ) F '( ) lim lim lim = = (By the FTC) 1 1 sin( ) = lim = 1

6 Question No. 6 of 1 Instruction: (1) Rea the problem statement an answer choices carefully () Work the problems on paper as neee (3) Question 6. The function t y = e t amits a point of inflection at Question #6 (A) since its erivative change sign at. (B) since its erivative remains strictly positive at. (C) is a critical point. (D) Its secon erivate changes sign at (E) None of these By the funamental Theorem of Calculus y is foun to be a strictly positive function. Incorrect justification. Zero is not a critical point of y. The secon erivative of y changes sign at. Therefore, is an inflection point. One of the given answer choices is correct. Please try again. The given function is twice ifferentiable. Thus to check whether a point is an inflection point, we must ascertain whether its secon erivative changes sign at that point. By the Funamental Theorem of -t Calculus, we fin that y' = e t e =. Therefore, by using the Chain Rule, y'' = e. Since the eponential remains positive, the sign of y epens solely on the sign of which is positive on the left of an negative on the right of. Hence y unergoes a sign change across. Whence is an inflection point.

7 Question No. 7 of 1 Instruction: (1) Rea the problem statement an answer choices carefully () Work the problems on paper as neee (3) Question 7. The eact value of ( + 1) is. 1 Question #7 (A) -1 (B) -56/15 (C) 56/ (D) (E) The integran remains positive from 1 o. Therefore the given efinite integral shoul be positive. It is the surface uner the curve escribe by the integran from 1 to. Use Linearity of integration together with the Funamental Theorem of calculus to compute the efinite integral. The integran remains positive from 1 o. Therefore the given efinite integral shoul be positive. It is the surface uner the curve escribe by the integran from 1 to. Use Linearity of integration together with the Funamental Theorem of calculus to compute the efinite integral. This is the only plausible answer. Use Linearity of integration together with the Funamental Theorem of calculus to confirm the answer. D. Incorrect! We epect a number as solution not a function. Use Linearity of integration together with the Funamental Theorem of calculus to compute the efinite integral. The integran remains positive from 1 o. Therefore the given efinite integral shoul be positive. It is the surface uner the curve escribe by the integran from 1 to. Use Linearity of integration together with the Funamental Theorem of calculus to compute the efinite integral. 1 ( + 1) = ( + 1) (Replacing the square root with a power.) = ( + ) (Using Algebra rules to epan the integran) 1 = (By the FTC an using the anti-erivation of the power functions) = = 56 15

8 Question No. 8 of 1 Instruction: (1) Rea the problem statement an answer choices carefully () Work the problems on paper as neee (3) - t Question8. Let F( ) = e e t. The equation of the tangent line to the graph of F at (,) is. Question #8 (A) y=1/ (B) y=-1 (C) y=- (D) y= (E) y= The given equation is not the equation of a line. A line trough the origin oes not have a non-zero -intercept. The slope is incorrect. Compute the slope of the tangent line by evaluating the erivative of the given function at. You will nee to apply both the prouct rule an the Funamental Theorem of Calculus to Calculate F (). Recall that the general equation of a line is y=m +b. From the previous calculation, you alreay know m. By forcing the line to pass trough (,), we get b=. The given equation is not the equation of a line. The equation of the tangent line to F through (,) is y = F() + F '()( ). From the formula of F, we get F()=. We are left with calculating F (). By applying both the prouct rule an the Funamental Theorem of Calculus, we obtain - t F '( ) = e e t + 1. Therefore F ()=1. Finally, the equation of the tangent line through (,) simplifies to y=.

9 Question No. 9 of 1 Instruction: (1) Rea the problem statement an answer choices carefully () Work the problems on paper as neee (3) Question 9. Given that ( ln ) = ln, evaluate the efinite integral ln. Question #9 (A) Does not eist. (B) - (C) 6 ln+ (D) 6 ln- (E) - Since the integran is continuous on [,], the given efinite integral is well-efine. Employ the premise of the problem to infer a particular anti-erivative of the integran an use the Funamental Theorem of Calculus to infer the correct answer. The natural logarithm is positive on [,]. Therefore the integral of the natural logarithm over [,] is epecte to be positive. Employ the premise of the problem to infer a particular anti-erivative of the integran an use the Funamental Theorem of Calculus to infer the correct answer. The natural logarithm is positive on [,]. Therefore the integral of the natural logarithm over [,] is epecte to be positive. The suggeste value is negative! Employ the premise of the problem to infer a particular anti-erivative of the integran an use the Funamental Theorem of Calculus to infer the correct answer. From the premise of the problem, we infer that a particular anti-erivative of ln is ln -. Use the Funamental Theorem of Calculus to conclue that the correct answer is 6 ln -. Since the integran is continuous on [,], the given efinite integral is well-efine. Employ the premise of the problem to infer a particular anti-erivative of the integran an use the Funamental Theorem of Calculus to infer the correct answer. The premise of the question says that Funamental Theorem of Calculus, ln = ln = ln (ln ) = 8ln ln + = 6 ln. ln is a particular anti-erivative of ln. Therefore, by the

10 Question No. 1 of 1 Instruction: (1) Rea the problem statement an answer choices carefully () Work the problems on paper as neee (3) 3 Question 1. Let y = cos( t ) t. Then y is. Question #1 (A) cos( ) (B) cos( t ) (C) 3 cos(9 t ) cos( t ) (D) 3 cos(9 ) cos( ) (E) cos(9 ) The given integral function is not in the form where the Funamental Theorem of calculus applies reaily. You have to split the formula of the function using a property of integration in such a way that the lower limits of integration in the resulting integrals are constant, rewrite the resulting integrals as composition of functions, then apply the linearity property of ifferentiation, the chain rule an the Funamental Theorem of Calculus to get y. You may also apply right away the formula obtaine in this tutorial following the route just escribe. The given function epens on not t! You have to split the formula of the function using appropriate properties of integration in such a way that the lower limits of integration in the resulting integrals are constant, rewrite the resulting integrals as composition of functions, then apply the linearity property of ifferentiation, the chain rule an the Funamental Theorem of Calculus to get y. You may also apply right away the formula obtaine in this tutorial following the route just escribe. The given function epens on not t! You have to split the formula of the function using appropriate properties of integration in such a way that the lower limits of integration in the resulting integrals are constant, rewrite the resulting integrals as composition of functions, then apply the linearity property of ifferentiation, the chain rule an the Funamental Theorem of Calculus to get y. You may also apply right away the formula obtaine in this tutorial following the route just escribe. Split the formula of the function using appropriate properties of integration in such a way that the lower limits of integration in the resulting integrals are constant, rewrite the resulting integrals as composition of functions, then apply the linearity property of ifferentiation, the chain rule an the Funamental Theorem of Calculus to get y. You may also apply right away the formula obtaine in this tutorial following the route just escribe. The given integral function is not in the form where the Funamental Theorem of calculus applies reaily. You have to split the formula of the function using a property of integration in such a way that the lower limits of integration in the resulting integrals are constant, rewrite the resulting integrals as composition of functions, then apply the linearity property of ifferentiation, the chain rule an the Funamental Theorem of Calculus to get y. You may also apply right away the formula obtaine in this tutorial following the route just escribe. v( ) Use the formula f() t t = u '( )(( f u )) + v '( )(( f v )). Inee the given integral function is in u ( ) v( ) the for f() t t with u()=, v()=3 an f( t) = cos( t ). Therefore, u ( ) y' = cos( ) + 3 cos[(3 ) ] = 3 cos(9 ) cos( ).