The Fundmentl Theorem of Clculus As we continue to study the re problem, let s think bck to wht we know bout computing res of regions enclosed by curves. If we wnt to find the re of the region below the curve f(x), we hve seen tht we cn think of the re of this region s limit of sums of res of pproximting rectngles, given by the formul A = lim n n f(x i ) x. i=1 In section 4.2, we defined the definite integrl of f from x = to x = b s b f(x) dx = lim n n f(x i ) x; we sw tht if f(x) on [, b], then the re of the region under f(x) from x = to x = b is precisely A = b i=1 f(x) dx. If, however, f(x) tkes on vlues less thn on the intervl from x = to x = b, then the integrl b f(x) dx describes the difference between res of regions bove the x-xis nd res of regions below the x-xis. In 4.2, we lso sw n exmple of the clcultion for b f(x) dx, which ws quite tedious. Our gol for this section is to find much quicker, less pinful wy to mke the sme clcultion. 1
As precursor to reching this gol, we begin by building new type of function one bsed on integrls. Below is the grph of function f(t): We re going to define the function g(x) bsed on f(t) using the formul f(t) dt. Notice tht choosing vlue for x is essentilly the sme s evluting the integrl f(t) dt with tht choice of x s the upper bound of integrtion; in other words, g is function so tht the number g(x) describes the re of the region below f(t) from to x (or the difference between res of regions bove nd below the x-xis). Exmple. Given the grph of f(t) below nd the function find 1. g(2) 2. g(4) 3. g(5) 4. g(6) 5. g(7) 6. g(1) 2
1. Since we know tht g(2) = 2 which is precisely the re of the region below f(t) from t = to t = 2, shded below: Since the shded region is tringle, its re is A = 1 2 b h = 1 2 2 4 = 4, so we know tht g(2) = 4. 3
2. Since g(4) = 4 we know tht g(4) is the re of the region shded below: We ve lredy found the re from t = to t = 2, so we need to focus on the re of the new region from t = 2 to t = 4; clerly, this is so 1 2 + 1 2 + 1 + 4 = 6, g(4) = 4 + 6 = 1. 3. Agin, we cn clculte g(5) = by finding the re of the shded region: 5 f(t) dt 4
The only new region is the one from t = 4 to t = 5, whose re is 2, so g(5) = 1 + 2 = 12. 4. To get g(6) = 6 we need to find the re of the shded region from t = to t = 6: Agin, the only new region is the one from t = 5 to t = 6; it s tringle, so its re is A = 1 2 1 2 = 1. Thus g(6) = 12 + 1 = 13. 5
5. Unfortuntely, g(7) = 7 f(t) dt is bit hrder to clculte since f(t) goes below the x-xis: We will need to clculte the re of the shded region below the x-xis, then subtrct it from the re of the region bove the x-xis. Agin, this new region is tringle, so it re is A = 1 2 1 2 = 1; but since the integrl trets this re s negtive, we know tht g(7) = 13 1 = 12. 6. Finlly, since g(1) = 1 we cn use the grph one more time to see tht the re of the only new region is A = 1 2 2 3 = 3 : 6
This region is below the x-xis, so gin the integrl trets it s negtive; we hve g(1) = 12 3 = 9. Since this new object g(x) is ctully function, it hs derivtive g (x) which we might wish to clculte but s of yet we hve no wy to do this. Fortuntely, the theorem below tells us exctly wht g (x) is: The Fundmentl Theorem of Clculus, Prt 1. If f(t) is continuous on [, b] then the function f(t) dt is s well, nd is differentible on (, b); the derivtive of g(x) is given by g (x) = f(x). In sense, the theorem tells us tht the derivtive of the integrl of f(x) is f(x). Exmple. Find g (x) given tht 2 4 sin t t 2 Since g(x) is defined s n integrl, we cn find its derivtive using the Fundmentl Theorem of Clculus: g (x) = 4 sin x x 2. dt. 7
Exmple. Given sin x t 3 dt, find g (x). Unfortuntely, we cn t directly pply the Fundmentl Theorem of Clculus here; g(x) is ctully composition of (inside) h(x) = sin x nd (outside) j(x) = t 3 dt. Thus we ll need to use the chin rule to integrte; fortuntely, we know the derivtives of both the inside nd outside functions: Using the chin rule, we hve h (x) = cos x nd j (x) = x 3. g (x) = h (x) j (h(x)) = (cos x) (j (sin x)) = (cos x) ((sin x) 3 ) = cos x sin 3 x. Thus g (x) = cos x sin 3 x. Finlly, we get to return to our originl question: How do we quickly evlute definite integrls such s The question is nswered by b f(x) dx? The Fundmentl Theorem of Clculus, Prt 2. If f(x) is continuous on [, b], nd if F (x) is ny ntiderivtive of f(x) (so tht F (x) = f(x)), then b f(x) dx = F (b) F (). FTC 2 sys tht, in order to evlute definite integrl of f(x), we merely need to evlute n ntiderivtive F (x) of f t the bounds b nd of integrtion! Tken together, the two prts of the Fundmentl Theorem of Clculus sy tht differentition nd integrtion re processes tht reverse ech other. 8
Exmple. Find the re A of the region below f(x) = x from x = to x = 1. In the previous section, we expressed this re s the definite integrl A = 1 x dx. Now we hve the tools to ctully clculte the integrl! FTC 2 tells us tht we should find ny ntiderivtive of f(x) = x = x 1/2 ; F (x) = 1 3 x 3/2 = 2 3 x3/2 2 will do. Since F (x) = 2/3x 3/2 is n ntiderivtive of f(x) = x, FTC 2 sys tht A = 1 x dx = F (1) F () = 2 3 (1)3/2 2 3 ()3/2 = 2 3 (1)3/2 = 2 3. Thus the re of the region below f(x) = x from x = to x = 1 is A = 2/3. Exmple. Evlute π cos x dx nd interpret your result. Agin, FTC 2 tells us tht evluting this integrl relly comes down to finding n ntiderivtive F (x) of the integrnd f(x) = cos x; F (x) = sin x will work. Thus π cos x dx = F (π) F () = sin π sin =. Since the vlue for the integrl is, we conclude tht the curve cos x encloses the sme re bove the x-xis s it does below the x-xis from x = to x = π; since the integrl counts re below the xis s negtive, the res cncel ech other, to yield n integrl of. This interprettion is confirmed by the grph of f(x) = cos x: 9
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