1 By Gökhan Bilhan Calculus 2 (Week 6)-Improper Integrals Part 1 How can we nd the area under the curve, or in other words, what is the result of the integral Z ∞ 1 lnx dx x2 Denition. Let's consider the integral ∞ Z f (x)dx. a Then we do: Z θ f (x)dx limθ7→∞ a If the case is Z b f (x)dx then we do: −∞ Z limθ7→−∞ b f (x)dx θ If the case is Z Z −∞ −∞ Z c ∞ f (x)dx = −∞ −∞ f (x)dx then we split the integral into two parts at any point c: Z ∞ f (x)dx + f (x)dx and then we go on. c Note that, if the result of the limit is a number then our integral's result is this number and we say that the integral is CONVERGENT. If the result of the limit is +∞ or −∞, then we say that the integral is DIVERGENT. Example: Now, let's calculate the integral Z 1 ∞ lnx dx x2 2 By Gökhan Bilhan .. Example: Z ∞ −∞ 1 dx 1 + x2 3 By Gökhan Bilhan Example: For what values of "p" does the integral Z 1 p>1 p<1 p=1 ∞ 1 dx converge, what is its value? xp 4 By Gökhan Bilhan Exercises 1. Determine whether the following integrals are convergent or not. In case an integral is convergent, evaluate it Z ∞ 1 Z ∞ ∞ 1 ∞ Z ∞ 1 dx x Z ∞ 1 dx Z dx 0,1001 1 √ dx x ∞ 1 Z 1 1 x1,0000000000001 1 1 x 1 Z 1 x1001 1 Z 1 dx x2 ∞ dx 1 dx xπ 1 √ dx x 1001 2. Evaluate the integral or state that it diverges. Z 0 ∞ (1 + x2 )(1 1 dx + arctanx) 5 By Gökhan Bilhan 3. Evaluate the integral or state that it diverges. Z 2 −∞ x2 2 dx +4 6 By Gökhan Bilhan PART 2 Denition. Let's consider the integral b Z f (x)dx. a θ. If at any of the boundaries, our function y = f (x) becomes not continuous, then we replace it with In particular, say f (a) is undened, i.e. at x = a, the function y = f (x) becomes discontinuous, then we do: Z b limθ7→a+ f (x)dx θ If say f (b) is undened, i.e. at x = b, the function y = f (x) becomes discontinuous, then we do: Z θ f (x)dx limθ7→b− a Example: Let's investigate the convergence of Z 0 1 1 dx 1−x 7 By Gökhan Bilhan Denition. Let's consider the integral Z b f (x)dx. a If at a point c among the boundaries a and b, our function y = f (x) becomes not continuous, then we split it into two parts at the point c and then apply the method described at the preceeding denition. i.e. Z b Z Z a b f (x)dx f (x)dx + f (x)dx = a c c Example: Let's investigate the convergence of Z 3 1 2 0 (x − 1) 3 dx 8 By Gökhan Bilhan Exercises 1. For which values of p, the integral Z 0 1 1 dx is convergent? xp 2. Evaluate the integral or state that it diverges. Z 1 3. Z 0 ∞ 1 dx x2 2 1 ds s s2 − 1 √ 9 By Gökhan Bilhan 4. Evaluate the integral or state that it diverges. Z 0 1 √ 4r dr 1 − r4 By Gökhan Bilhan Various Exercises 1. 2. 1 1 −1 x3 Z 2 Z 2 s+1 √ ds 4 − s2 4 1 p dx |x| 0 3. Z −1 4. Z dx π 2 cotxdx 0 10 11 By Gökhan Bilhan 5. Z ∞ √ 4 6. Z ∞ √ 1 7. Z 2 ∞ √ 1 dx x−1 x+1 dx x2 1 x2 − 1 dx