Mth 126 Clculus Sec. 5.6: Substitution nd Are Between Curves I. U-Substitution for Definite Integrls A. Th m 6-Substitution in Definite Integrls: If g (x) is continuous on [,b] nd f is continuous on the rnge of u=g(x), then B. Exmples 1. cos 3 (x) sin(x) dx b ( ) f g ( x ) g' ( x) dx = f( u) du gb ( ) g ( ) 4 2. 3y + 4 dy 6 3. tn() d
3 4. sin 2 (3) d NOTE: To evlute cos 2 (x) = 1 + cos(2x) 2 1 cos(2x) sin 2 (x)dx nd cos 2 (x)dx use the trig identities sin 2 (x) = 2 nd then use substitution rule. nd 5. ( 2 + tn( 2 ))sec 2 ( 2 )d 2
6. 1 1 2 1 t 2 ( 2 + 1 t ) 3 dt 7. (3+ xe x2 )dx ln 2 8. 9 dt = t 3 + t 4 ( ) 3
4 3 9. 7x x 4 dx 4 II. Symmetry A. Integrls of Symmetric Functions: Suppose f is continuous on [-,]. () If f is even, i.e., f(-x)=f(x), then f( x) dx = 2 f( x) dx. (b) If f is odd, i.e., f(-x)=-f(x), then f( x) dx =. B. Exmples 3 ( ) 2 dx 1. x -2 1 5 2. x 25 x 2 dx 5
III. Overview of Are Between Two Curves With few modifictions the re under curve represented by definite integrl cn be extended to find the re between to curves. Observe the following grphs of f(x) nd g(x). Since both grphs lie bove the x - xis, we cn geometriclly interpret the re of the region between the grphs s the re of the region under f minus the re of the region under g. Although we hve just shown one cse, i.e. when f nd g re both positive, this condition is not necessry. All we need in order to evlute the integrnd [f(x) - g(x)] is tht f nd g both be continuous nd g(x)< f(x) on the intervl [,b]. IV. Are of Region Between Two Curves Integrting with respect to x A. Finding the Are Between the Curves Using Verticl Rectngles; Integrting wrt x. 1. If f nd g re continuous on [,b] nd g(x) < f(x) for ll x in [,b], then the re of the region bounded by the grphs of f nd g nd the verticl lines x= nd x=b is b A = [ f( x) gx ( ) ] dx ; where f(x) is the upper curve & g(x) is the lower curve. In the next picture, you cn see tht we simply prtition the intervl [,b] nd drw our representtive rectngles. I've drwn few rectngles to illustrte the concept.
B. Steps to Find the Are Between Two Curves (wrt x) 1. Grph the curves nd drw representtive rectngle. This revels which curve is f (upper curve) nd which is g (lower curve). It lso helps find the limits of integrtion if you do not lredy know them. 2. Find the limits of integrtion; you my need to find the pts of intersection. 3. Write formul for f(x) - g(x). Simplify it if you cn. (c) Integrte [f(x) g(x)] from to b. The number you get is the re. Note: You cn use grphing clcultor, Mtlb or Mthemtic to drw the functions, this will gretly speed up your work. The min reson for doing step 1 is to see how mny times the grphs cross (if ever) in the specified domin. It lso is to determine which function is dominnt in which intervls. If you don't hve quick mens to grph the functions, you cn set the two functions equl to ech other nd try to solve for where they re equl. This will give you the intersection points of the two grphs. Once you know these, you cn determine which function is dominnt in which subintervls. Then just proceed s before. You cn use Mthemtic's Solve or FindRoot commnds to help in finding intersection points. C. Exmples 1. Find the re of the region bounded by y = x + 2 nd y = x 2.
2. Find the re of the region below tht is bounded by y = 1 x 2, y = (x 1) 2 nd y = 1. y = 1 y = 1 x 2 y = (x 1) 2 3. Find the re of the region bounded by y = x 3, y = 2 x 2, x =, nd x = 2.
V. Are of Region Between Two Curves - Integrting with respect to y. Wht is relly net is tht insted of integrting with respect to x, you cn lso integrte with respect to y, (your representtive rectngles will now be prllel to the x - xis insted of the y - xis). For Exmple, suppose we hd the following grphs: In the first picture, we only need to integrte one time. In the second picture, we would hve to integrte twice using different limits of integrtion nd different functions (this is symbolized by the different color rectngles tht you would hve to construct). A. Finding the Are Between the Curves Using Horizontl Rectngles; Integrting wrt y. If r nd l re continuous on [c,d] nd l(y) < r(y) for ll y in [c,d], then the re of the region bounded by the grphs of r nd l nd the horizontl lines y=c nd y=d is d A = r( y) l( y) dy ; where r(y) is the right curve & l(y) is the left curve. c B. Steps to Find the Are Between Two Curves 1. Grph the curves nd drw representtive rectngle. This revels which curve is r (right curve) nd which is l (left curve). It lso helps find the limits of integrtion if you do not lredy know them. 2. Find the limits of integrtion; you my need to find the pts of intersection. 3. Write formul for r(y)-l(y). Simplify it if you cn. 4. Integrte [r(y)-l(y)] from c to d. The number you get is the re. C. Exmples 1. Find the re of the region bounded by x = 4 y 2 nd x = 2 + y.
2. Find the re of the region below tht is bounded by y = 1 x 2, y = (x 1) 2 nd y = 1. y = 1 y = 1 x 2 y = (x 1) 2