Math 181, Exam 1, Spring 2013 Problem 1 Solution. arctan xdx.

Transcription

1 Math, Exam, Sring 03 Problem Solution. Comute the integrals xe 4x and arctan x. Solution: We comute the first integral using Integration by Parts. The following table summarizes the elements that make u the technique. u x du dv e 4x v 4 e4x Using the Integration by Parts formula we have udv uv vdu xe 4x 4 xe4x e 4x 4 xe 4x 4 xe4x 6 e4x + C The second integral is comuted by Integration by Parts as well. The following table summaries the elements that make u the technique. u arctan(x) du x + dv v x Using the Integration by Parts formula we have udv uv vdu arctan(x) x arctan(x) arctan(x) x arctan(x) x x + ln(x +)+C

2 . Comute the integral Math, Exam, Sring 03 Problem Solution x x +4. Solution: Wecomutetheintegralusingthetrigonometricsubstitutionx tan, sec d. After substituting these exressions into the integral and simlifying we obtain: x x +4 sec d ( tan ) ( tan ) +4 sec 4tan 4tan +4 d sec 4tan 4(tan +) d sec 4tan 4sec d sec 4tan sec d sec 4 tan d To evaluate the resulting trigonometric integral we rewrite the integrand in terms of sin and cos by using the definitions The integral then transforms into We can either sec cos, sec tan d sin tan cos cos sin d () rewrite the integrand as cot csc and use the fact that this is the derivative of csc or () use the substitution u sin, du cos d. In either case we obtain the result: cos sin d Therefore, our integral takes the form x x +4 csc + C 4 csc + C

3 We must finish the roblem by writing csc in terms of x. Sinceweknowthatx tan we have tan x OPP ADJ where OPP and ADJ are the oosite and adjacent sides of a right triangle, resectively, where oosite refers to the length of the side across from. Using the Pythagorean Theorem, the hyotenuse of this triangle is x +4. Therefore, csc sin HYP OPP x +4 After substituting this exression into our result we find that x x x +4 x +4 + C 4x

4 Math, Exam, Sring 03 Problem 3 Solution 3. The region between the x-axis and the arabola y x is rotated about the line x. Find the volume of the resulting solid. Solution: Theregionbeingrotatedislottedbelow. In this case, we use the Shell Method because the integration with resect to x is easier to erform. The volume formula is V b a (radius)(height) The height of each shell is given by x. Since the region is being rotated about the axis x,theradiusofeachshellisgivenby x. The interval over which the integral will take lace is x tox sincethesearetheointswherethearabolay x intersects the x-axis. Therefore, the volume of the solid is V V V V V ale x ( x)( x ) ( x x + x 3 ) ( x ) x 3 3 0

5 Math, Exam, Sring 03 Problem 4 Solution 4. Comute the integrals x x and x +x +5. Solution: The integrand of is a rational function whose denominator factors x x into x(x ). Thus, we will use the method of artial fractions. The artial fraction decomosition of the integrand is After clearing denominators we find that x(x ) A x + B x A(x ) + Bx When x 0wehaveA the integral as follows: andwhenx wehaveb. Therefore,wemayevaluate x x x + x x x ln x +ln x + C The integrand of the integral is a rational function but the denominator x +x +5 is an irreducible quadratic. Therefore we begin by comleting the square: x +x +5(x +x +)+5 (x +) +4 At the same time we can rewrite the numerator as (x +)+. Theintegralcan then be slit into the sum of two integrals x +x +5 x + (x +) +4 + (x +) +4 Letting u x +,du we obtain: x +x +5 u u +4 du + u +4 du The first integral on the right hand side may be evaluated using the substitution v u +4, dv udu and the second integral may be evaluated using the trigonometric substitution

6 u tan, du sec d.thesumoftheseintegralstransformsandevaluatesasfollows: x +x +5 u u +4 + u +4 du dv v + d ln v + + C u ln(u +4)+arctan + C where we used the fact that v u +4and arctan( u )towriteouranswerintermsof u. Wemusttakeitastefurtherandwriteouranswerintermsofx. Weusethefactthat u x +toobtain: x +x +5 ln((x +) +4)+arctan x +x +5 ln(x +x +5)+arctan x + + C x + + C

7 Math, Exam, Sring 03 Problem 5 Solution 5. Find the arc length of the grah of f(x) lnx Solution: The arc length formula we will use is x from to e. L b a +f 0 (x) The derivative f 0 (x) is f 0 (x) x x 4 Uon adding to the square of f 0 (x) wefindthattheresultisaerfectsquare.thedetails are outlined below: +f 0 (x) x + x 4 +f 0 (x) + x + x 6 Therefore, the arc length is +f 0 (x) x + + x 6 +f 0 (x) x + x 4 e L +f0 (x) e L x + x 4 L aleln(x)+ x e L aleln(e)+ aleln() e + L + (e )