1.8. Integration using Tables and CAS

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1 1.. INTEGRATION USING TABLES AND CAS Integration using Tables and CAS The use of tables of integrals and Computer Algebra Systems allow us to find integrals very quickly without having to perform all the steps for their computation. However we often need to modify slightly the original integral and perhaps complete or simplify the answer. Example: Find the following integral using the tables at the end of Steward s book: x 1 dx =. x Answer: In the tables we find the following formula No. 41: a u du = a u u a cos 1 a u + C, hence, letting a = 1, u = 1 we get the answer: 1 x dx = 1 x x cos 1 1 x + C. Example: Find the integral: x dx = 9 + 4x Answer: In the tables the formula that resembles this integral most is No. 6: u du a + u = u ( a + u a ln u + ) a + u + C, hence making a = 3, u = x: x dx = 1 u du 9 + 4x a + u = 1 { u ( a + u a ln u + ) } a + u + C = x 9 + 4x 9 (x 16 ln + ) 9 + 4x + C. Example: Find the same integral using Maple.

2 1.. INTEGRATION USING TABLES AND CAS 40 Answer: In Maple we enter at the prompt: > int(x^/sqrt(9+4*x^),x); and it returns: x 9 + 4x 9 ( ) 16 arcsinh 3 x First we notice that the answer omits the constant C. On the other hand, it involves an inverse hyperbolic function: arcsinh x = ln (x + ) 1 + x, hence the answer provided by Maple is: ( x 9 + 4x 9 ) 16 ln x x = 9 x 9 + 4x 9 (x 16 ln + ) 9 + 4x ln(3), so it differs from the answer found using the tables in a constant 9 ln(3) 3 which can be absorbed into the constant of integration.

3 1.9. NUMERICAL INTEGRATION Numerical Integration Sometimes the integral of a function cannot be expressed with elementary functions, i.e., polynomial, trigonometric, exponential, logarithmic, or a suitable combination of these. However, in those cases we still can find an approximate value for the integral of a function on an interval Trapezoidal Approximation. A first attempt to approximate the value of an integral b f(x) dx is to compute its Riemann a sum: n R = f(x i ) x. i=1 Where x = x i x i 1 = (b a)/n and x i is some point in the interval [x i 1, x i ]. If we choose the left endpoints of each interval, we get the left-endpoint approximation: n L n = f(x i 1 )) x = ( x){f(x 0 ) + f(x 1 ) + + f(x n 1 )}, i=1 Similarly, by choosing the right endpoints of each interval we get the right-endpoint approximation: n R n = f(x i ) x = ( x){f(x 1 ) + f(x ) + + f(x n )}. i=1 The trapezoidal approximation is the average of L n and R n : T n = 1 (L n+r n ) = x {f(x 0)+f(x 1 )+f(x )+ +f(x n 1 )+f(x n )}. Example: Approximate 1 0 x dx with trapezoidal approximation using 4 intervals. Solution: We have x = 1/4 = 0.5. The values for x i and f(x i ) = x i can be tabulated in the following way: i x i f(x i )

4 1.9. NUMERICAL INTEGRATION 4 Hence: L 4 = 0.5 ( ) = , R 4 = 0.5 ( ) = So: T 4 = 1 (L 4 + R 4 ) = 1 ( ) = Compare to the exact value of the integral, which is 1/3 = Midpoint Approximation. Alternatively, in the Riemann sum we can use the middle point x i = (x i 1 + x i )/ of each interval [x i 1, x i ]. Then the midpoint approximation of b f(x) dx is a M n = ( x){f(x 1 ) + f(x ) + + f(x n )}. Example: Approximate 1 0 x dx with midpoint approximation using 4 intervals. Solution: We have: Hence: i x i f(x i ) M 4 = 0.5 ( ) = Simpson s Approximation. Simpson s approximation is a weighted average of the trapezoidal and midpoint approximations associated to the intervals [x 0, x ], [x, x 4 ],..., [x n, x n ] (of length

5 1.9. NUMERICAL INTEGRATION 43 x each): S n = 1 3 (M n + T n ) = 1 [ ( x){f(x 1 ) + f(x 3 ) + + f(x n 1 )} 3 + x {f(x 0) + f(x ) + f(x 4 ) + + f(x n ) + f(x n )} = x 3 {f(x 0) + 4f(x 1 ) + f(x ) + 4f(x 3 ) + f(x 4 ) + + f(x n ) + 4f(x n 1 ) + f(x n )}. Example: Approximate 1 0 x dx with Simpson s approximation using intervals. Solution: We use the previous results and get: S = 1 3 (M 4 + T 4 ) = 1 ( ) = 1/3. 3 Note: in this particular case Simpson s approximation gives the exact value in general it just gives a good approximation Error Bounds. Here we give a way to estimate the error or difference E between the actual value of an integral and the value obtained using a numerical approximation Error Bound for the Trapezoidal Approximation. Suppose f (x) K for a x b. Then the error E T in the trapezoidal approximation verifies: ] E T K(b a)3 1n Error Bound for the Midpoint Approximation. Suppose f (x) K for a x b. Then the error E M in the trapezoidal approximation verifies: K(b a)3 E M. 4n Error Bound for the Simpson s Rule. Suppose f (4) (x) K for a x b. Then the error E S in the Simpson s rule verifies: E S K(b a)5 10n 4.

6 1.9. NUMERICAL INTEGRATION 44 Example: Approximate the value of π using the trapezoidal, midpoint and Simpson s approximations of for n = 4. Estimate the error. Answer: First note that: x dx x dx = 4 [ tan 1 x ] 1 0 = 4π 4 = π, so by approximating the given integral we are in fact finding approximated values for π. Now we find the requested approximations: (1) Trapezoidal approximation: T 4 = 1/4 {f(0) + f(1/4) + f(1/) + f(3/4) + f(1)} = For estimating the error we need the second derivative of f(x) = 4/(1 + x ), which is f (x) = (3x 1)/(1 + x ) 3 so we have f (x) = 3x 1 (3x + 1) 1 + x 3 (1 + x ) 3 for 0 x 1, hence ( ) 1 = 3 3 (1 0)3 E T = () Midpoint approximation: M 4 = 1 {f(1/) + f(3/) + f(5/) + f(7/)} 4 = The error estimate is: E M 3 (1 0)3 4 4 =

7 (3) Simpson s rule: 1.9. NUMERICAL INTEGRATION 45 S 4 = 1/4 {f(0) + 4f(1/4) + f(1/) + 4f(3/4) + f(1)} 3 = For the error estimate we now need the fourth derivative: f (4) (x) = 96(5x 4 10x + 1)/(1 + x ) 5, so f (4) 96( ) (x) = for 0 x 1. Hence the error estimate is 1536 (1 0)5 E S =