MATH260Week 6 Lab Name: Antiderivatives According to the first part of the fundamental theorem of calculus, the antiderivative reverses the derivative. If f(x) is a derivative, F(x) is the antiderivative. Directions: Look at the examples below and answer questions 1 and 2. Let f(x) be a derivative and F(x) be the anitderivative. a.) f(x) = 3x2 2 +1 3x F( x )= =x 3 2+ 1 b.) f(x) = 5x - 6 3 c.) 1 +1 F( x )= 0+1 5x 6x 1+ 1 1 5 = x 26 x 2 f (x)= x 1 3 + 4 +1 1 x4 4 x3 3 x 3 1 F( x )= = x3+ 3 4 3 4 3x 3 2) Find the antiderivative of f(x) = 3x5 + x - 4. Show all work. 3) How is an antiderivative related to a derivative? How can that relationship help you to check your antiderivatives and integrals answers? 4) What do the derivatives of the following antiderivatives have in common? F(x) = 3x2 + 5 , G(x) = 3x2 + 9 , H(x) = 3x2 - 11 Indefinite Integrals: f (x) dx=F (x)+C Used for finding the general form of the antiderivative. --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- Constant of Integration: C is the constant of integration that must be attached to an indefinite integral. It represents any of the constants that could be part of F(x). Directions: Look at the examples, then find each of the integrals below. 4 x3 6 x dx 3 x+ 5 dx 3 2 x +5 x +C F(x) = 2 F(x) = 2 3 3 2 2 x 2 +C x F(x) = 6) Find the integral for f (x)= x + 2 4 x5 dx 3 x+ x 2 dx F(x) = 4 x 3 x + C 4 +7 x1 2 . Show all work. x 4 x5 3+ C 1 12 Integration by Substitution If part of the function is a derivative of the other part of the function, then integration by substitution can be applied. If the form is true, the following method will apply. 1) 2) 3) 4) Label part of the function u and then find du, the derivative of u. Substitute u and du into the integral to replace all expressions in x. Integrate the substituted function. Replace u with the original expression. ------------------------------------------------------------------------------------------------------------------------------------------------------------------------- Examine each example below and answer questions 7 and 8. Integral: Let u = 2 2 x ( x +4 ) dx 2 Integral: 2 x + 4 , then du = 2x dx x3 4 5 dx ( 5 x +9 ) 4 Let u = 5 x +9 , then du = 1 du=x3 dx 20 and Substitution: u2 du Substitution: x3 1 4 5 dx= 20 du u5 ( 5 x +9 ) Solution: and 1 F ( u )= u 3 3 3 1 F ( x )= ( x 2 +4 ) +C 3 7) How is u determined? How is du determined? 8) Why is there a 1/20 in the substitution for ? Solution: and F ( x )= F ( u )= 1 4 3 20 x dx 4 80 ( 5 x + 9 ) 1 1 [ 20 4 u4 +C 9) Identify u, du, the substituted integral, the correct integration, and the final answer. 5 3 x 2 ( x 32 ) dx Final Answer = 10) Find 3 x 4+ 3 9 dx 4 x3 . Show u and du, the substituted integral, and the final answer. 11) Find x4 dx 3 . Show u and du, the substituted integral, and the final answer. ( 7 x 5+1 ) Finding C: You have already seen that a general solution to an integral can be found by performing antidifferentiation. However, because the derivative of a constant is zero, any integral has many solutions, each differing from the others by C, a constant of integration. In many applications of integration, you are given enough information to determine a particular solution (that is, you can find C). To do this, all you need is an ordered pair from the original function. 12) If s'(t) = v(t), and s''(t) = a(t), use the fact that integration reverses differentiation (+ C) and the given information about s(t) to find the particular solution to the following problem. If a model rocket's velocity is given by a(t)= -32 m/s 2 and v(0)= 10 and position at 1 second is 1,000 m, find s(t). Hint: Integrate a(t) to find v(t), then integrate v(t) to find s(t), finding each C using the given info. Part II: Finding the Area Under a Curve The area under a curve can be found by filling the area with rectangles, then finding the area of each rectangle, then adding the areas together. To find an approximate area under a curve, find the total area for both left and right rectangles, then average the two. These are called Riemann sums and are the basis of the integral. --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- 13) For the function f(x) = (x - 2)2 + 2 estimate the area under the curve on the interval [0, 4] using 5 left rectangles, then 5 right rectangles, then find the average of the two areas. Finally, write and find the exact area by using a definite integral. Give answers in exact form (no decimals). Sketch five left rectangles: Based on your sketch, is this estimate an over- or underestimate? Why? Fill in the blanks below. What is the width of each rectangle? Width = x1 = f(x1) = x2 = 4 =.8 5 x3 = 4 4 8 + = =1.6 5 5 5 f(x3) = x4 = 8 4 12 + = =2.4 5 5 5 f(x4) = x5 = 12 4 16 + = =3.2 5 5 5 f(x5) = Total area of the left rectangles = f(x2) = f ( 4 )= 86 =3.44 5 25 f ( 8 )= 54 =2.16 5 25 f ( 12 )= 54 =2.16 5 25 f ( 16 )= 86 =3.44 5 25 4 86 54 54 86 344 6+ + + + = =13.76 5 25 25 25 25 25 ( ) Sketch five right rectangles: 4 Based on your sketch, is this estimate an over- or underestimate? Why? 4 5 x1 = 4 =.8 5 f(x1) = x2 = 4 4 8 + = =1.6 5 5 5 f(x2) = x3 = 8 4 12 + = =2.4 5 5 5 f(x3) = x4 = 12 4 16 + = =3.2 5 5 5 f(x4) = x5 = 16 4 20 + = =4 5 5 5 f(x5) = Total area of the left rectangles = Average of the left & right rectangles = f ( 4 )= 86 =3.44 5 25 f ( 8 )= 54 =2.16 5 25 f ( 8 )= 54 =2.16 5 25 f ( 16 )= 86 =3.44 5 25 f ( 5 ) =11 4 86 54 54 86 324 + + + +5 = =12.96 5 25 25 25 25 25 ( ) 669 =13.38 25 The Definite Integral and Area Under a Curve If you use the above method and let the number of rectangles be infinite (the limit), the area can be found to a very accurate measure. To find the sum of the area of an infinite number of rectangles, use the second part of the fundamental theorem of calculus. b f ( x)dx=F (b)F (a) a 14) Using the fundamental theorem of calculus above, write the correct integral and find the exact area for problem 13 above. Some Basic Rules of the Definite Integral: all f(x) must exist and be continuous on [a, b]. a 1) f ( x) dx a =0 (if the width is zero, there can be no area) b 2) f ( x) dx a a b 3) f ( x) dx = - (if you switch the limits of integration, switch the sign) b c f ( x) dx = a b f (x)dx a + f (x) dx c , for a < c < b (If c is between a and b, then you can break the integral in two at c and add them together to get the total area; i.e., the area from 2 to 5 is equal to the area from 2 to 3 plus 3 to 5.) b 4) kf ( x)dx a b = k f ( x) dx a (you can move any constant factor of the function outside of the integral, then integrate and multiply it back over the answer.) b 5) [f (x) g (x)]dx a b = b f (x) dx g( x ) dx a a (the integral of a sum or difference is the sum or difference of the integrals) Note: there is no such rule for multiplication and division. 15) Using the rules for definite integrals above, answer the following questions a)-d). 0 a) Why can't you find the value of this definite integral? x1 dx 2 3 b) Why can't you find the value of this definite integral? x 2+ 3 x +1 dx 3 3 c) Will the following help to find the value of this definite integral? 3 3 x (x+ 3)dx= x 2 1 1 x 2 ( x+ 3) dx 1 3 2 dx ( x+3) dx 1 Why or why not? d) How is it possible to get a negative area? 16) Find the area under the curve 3 f ( x )=x 9 x from [-3, 3] by using the fundamental rule of b calculus: f ( x)dx=F (b)F (a) a . Show all work. 17) Below is the graph of f(x) = x3 - 9x with the area between the graph and the x-axis from -3 to 3 shaded. How does it help to explain your answer to 16) above? How would you find the actual area? 18) The derivative gives an expression for the slope of a curve and the definite integral gives the area under the curve, two very different ideas. Name two things the derivative and the integral have in common. MATRH260Week 7 Lab Name: Logarithmic Integrals Examine each example below then answer questions 1-3. Integral: 1 x dx Solution: ln |x| Integral: Integral: Solution: 2x x2 + 4 dx Solution: ln |x 2+4|+C ln |x +3|+C Integral: +C 1) What is the formula for finding 1 x3 dx 2 x3 dx Solution: 1 1 ln |2 x 3|+C 2 du ? u 2) How is substitution being used to solve examples and ? Show u and du. 3) Why can't we find the integral 1 x dx using the following set-up? 1 x dx= x1 dx 4) Fill in the blanks below with the correct information for the following integral: u= Substitution = du = Integration = Final answer = u= Substitution = du = Integration = Final answer = u= Substitution = Final answer = du = Integration = 3 4 x +5 dx Exponential Integrals Examine each example below and answer questions 7-10. e x dx Integral: x Solution: e +C e 3 x dx Integral: Solution: Solution: 2 e x +C sec 2 (2 x )etan ( 2 x) dx Integral: 1 3x e +C 3 7) What is the formula for finding 2 2 xe x dx Integral: Solution: e u du 1 tan (2 x) e +C 2 ? 8) How is substitution being used to help solve examples , , and ? Show u, du, and the substituted integral for each. 9) Fill in the blanks below with the correct information for the following integral: u= du = Substitution = Integration = Final answer = 10) Fill in the blanks below with the correct information: u= Substitution = Integration = Final answer = du = se c 2 (8 x)e 3 tan (8 x) dx 3 5 x 2 e x dx Bandwidth of a Series Resonance Circuit The area under the curve shows all acceptable signals with frequency greater than f L and less than f H that can pass through the resonance circuit. This is used in radio receivers to tune for different channels. 11) If the following equation I (f ) represents the curve for a frequency bandwidth similar to the one above, find the area under the curve from f L= .8 H to f H. = 1.2 H. Round all values to the 100ths. f 1 2 .2 2 , xmin= -2, xmax = 3, ymin= -2, ymax = 3 2 1 I ( f )= e .2 2 Trigonometric Integrals Some of the trig integrals are the reverse of their derivatives. cos x se c 2 x sin x dx = sin x + C cs c 2 x dx = tan x + C sec x tan x dx = -cos x + C dx = -cot x + C csc x cot x dx dx = sec x + C = -csc x + C And some require reciprocal identities and integrate to natural logs. sec x dx = ln |sec x + tan x| + C cot x dx csc x dx = -ln |csc x + cot x| + C tan x dx = ln|sin x| + C = -ln|cos x| + C All should be memorized; remember that you can always check an integrand by differentiating. Sine and Cosine Integral: Solution: Integral: sin ( 2 x ) dx Integral: 1 cos (2 x ) +C 2 Solution: x sin ( x 2 +5 ) dx Integral: 12) Use u-substitution and the above examples to find 13) Use u-substitution to find 3 x cos ( x2 +8 ) dx cos ( 4 x ) dx 1 sin ( 4 x ) +C 4 9 x 2 cos ( x3 15 ) dx e 2 x sin ( e 2 x ) dx Secant and Cosecant, Tangent and Cotangent Examine each example below and answer the following questions. Integral i. Solution 1 ln |sec (6 x ) + tan ( 6 x )|+C 6 sec ( 6 x ) dx ii. x 4 csc ( x 57 ) dx iii 1 x 2 e 1 x 2 1 x 2 sec ( e ) tan ( e ) dx 1 ln|csc ( x 57 ) +cot ( x 57 )|+C 5 1 x 2 2 sec ( e )+ C iv. csc ( 10 x ) cot ( 10 x ) dx 1 csc ( 10 x )+ C 10 v. cot ( 10 x ) dx 1 ln |sin ( 10 x)|+C 10 vi. 2 x tan ( 10 x 2 ) dx 1 ln cos ( 10 x 2 )+C 10 *Note: absolute value symbols ensure that the argument for the natural logarithm is always positive and the result is real. 14) Use differentiation to show that each of the solutions for the integrals i. , iii., v., vi. above is true. 15) If cot ( u ) du=ln|sin u|+C what does cot ( 1 x ) dx 2 = . 4 xmin = -1, x-max = 5, y min=-3, y-max = 2, radian mode. 16) Graph the function sin(ex) and find the area from 0 to 17) 3 x 5 csc ( x 6 ) dx 18) e3 x sec ( e3 x) tan ( e3 x ) dx 19) Break the integral into three less complicated integrals, then integrate. a+b a b = + Hint: rewrite using the idea that , reduce, rewrite using reciprocal identities, c c c then integrate. si n2 ( x )+ sin ( x)+cos( x) dx si n3 ( x )