PRACTICE PROBLEMS: SET 7 MATH 101: PROF. DRAGOS GHIOCA 1. Problems Problem 1. Determine with proof whether the following integrals are divergent or convergent. (a) ∫ ∞ 2 x +2 dx x4 + x 2 (b) ∫ ∞ x dx 2+1 x 1 (c) ∫ ∞ sin(x) dx 3+1 x 0 Problem 2. Determine with proof whether the following integral is divergent or convergent. If it is convergent, compute its value. ∫ 3 2 (x − 1)− 5 dx 1 1 2 MATH 101: PROF. DRAGOS GHIOCA 2. Solutions Problem 1. In each case we first guess what is the correct answer, and then try to replace the function we integrate with a simpler function which would be useful for us to prove our guess is indeed correct. (a) The order of magnitude for the given fraction is x12 (by taking the∫quotient ∞ of the leading terms of both numerator and denominator). Since 2 x12 dx is convergent, we suspect the integral is convergent; so we prove our guess is indeed correct. In order to do this we have to compare our function with a larger function whose integral would still be convergent. We note that for x ≥ 2 we have x4 + x > x4 and thus x2 + 2 x2 + 2 1 2 < = 2 + 4. 4 4 x +x x x x Therefore ∫ ∞ 2 x2 + 2 dx x4 + x ∫ < = = = = = = ∞ 1 2 + 4 dx 2 x x ∫2 ∞ ∫ ∞ 1 2 dx + dx 2 4 x x 2 2 ∫ t ∫ t 1 2 lim dx + lim dx t→∞ 2 x2 t→∞ 2 x4 ( ) ( ) 1 t 2 lim − | + lim − 3 |t2 t→∞ x 2 t→∞ 3x ) ) ( ( 2 1 1 1 lim − + lim − 3 t→∞ 2 t→∞ 12 t 3t 1 1 + 2 12 7 , 12 which proves that indeed the given integral is convergent. Note that the original function we integrated was positive and therefore we were allowed to use the usual comparative test for convergence/divergence of integrals. (b) The order of magnitude of the given fraction is x1 , whose integral on the interval [1, ∞] is divergent. So, we will compare our function to a suitable function which would prove the divergence of our integral. We need this time to choose a function smaller than x2x+1 ; so x1 is not the right choice since x 1 < . 2 x +1 x However we note that x2 ≥ 1 on the interval [1, ∞]; PRACTICE PROBLEMS: SET 7 3 therefore x2 x x 1 ≥ 2 = . 2 +1 x +x 2x Then ∫ ∞ 1 x2 x dx +1 ∫ ∞ 1 dx 2x 1 ∫ t 1 = lim dx t→∞ 1 2x ( ) ln(x) t = lim |1 t→∞ 2 ln(t) = lim , t→∞ 2 ≥ which diverges to infinity, thus proving that indeed our integral is divergent. Note that the original function we integrated was positive and therefore we were allowed to use the usual comparative test for convergence/divergence of integrals. (c) We know that | sin(x)| ≤ 1 for all x. Thus the integral behaves roughly as ∫ 0 ∞ 1 dx x3 + 1 which is convergent. However, in order to use the comparative test for convergence/divergence of integrals we have to take absolute value of the integral and only then apply our test. In this problem there is also an extra complication since we cannot simply drop the +1 from the denominator since then we would have a problem integrating near x = 0. Therefore we will split the integral as follows. (Note that we need to take first absolute 4 MATH 101: PROF. DRAGOS GHIOCA values since we want to exploit the inequality | sin(x)| ≤ 1.) ∫ ∞ sin(x) dx x3 + 1 0 ∫ ∞ sin(x) ≤ x3 + 1 dx 0 ∫ ∞ 1 ≤ dx 3+1 x 0 ∫ 1 ∫ ∞ 1 1 = dx + dx 3+1 3+1 x x 0 1 ∫ 1 ∫ ∞ 1 1 < dx + dx 3 1 x 0 1 ∫ t 1 = 1 + lim dx t→∞ 1 x3 ( ) 1 = 1 + lim − 2 |t1 t→∞ 2x ) ( 1 1 = 1 + lim − 2 t→∞ 2 2t 1 = 1+ 2 3 = ; 2 therefore the integral is indeed convergent. Note that in the above estimations we used that 1 ≤ 1 + x3 on [0, 1] and x3 < x3 + 1 on [1, ∞]. Problem 2. Using the easy substitution u = x − 1, we immediately compute ∫ 3 2 5 (x − 1)− 5 dx = (x − 1) 5 + C . 3 Therefore, we compute ∫ 2 (x − 1)− 5 dx ∫ 3 2 = lim (x − 1)− 5 dx t→1 t ( ) 3 5 5 (x − 1) |3t = lim t→1 3 3 3 5 · 25 5 − · (t − 1) 5 t→1 3 3 3 5 · 25 . = 3 = lim This proves the integral is actually convergent and it equals 5 √ 5 8 3 .