Algebra Integral calculus, Exercises of Mathematics

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ALGEBRA

Integral Calculus

Exercises

6.1 Antidifferentiation.

The Indefinite Integral

In problems 1 through 7, find the indicated integral.

1. R √ xdx Solution.

5. ex^^6 dx

Solution. Let P ( t ) denote the population of the community t years from now. Then the rate of change of the population concerning time is the derivative . It follows that the population function P ( t ) is an antiderivative of 0_._ 6 t^2 + 0_._ 2 t + 0_._ 5. That is, P ( t ) = Z P 0 ( t ) dt = Z (0_._ 6 t^2 + 0_._ 2 t + 0_._ 5) dt = = 0_._ 2 t^3 + 0_._ 1 t^2 + 0_._ 5 t + C for some constant C. During the next 2 years, the population will grow on behalf of P (2) C = Hence, the pollution in the lake will increase on behalf of 5 · 3 = 15 units.

10. An object is moving so that its speed after t minutes is v ( t ) = 1+4 t +3 t^2 meters per minute. How far does the object travel during 3 rd minute? Solution. Let s ( t ) denote the displacement of the car after t minutes. Since it follows that s ( t ) = Z v ( t ) dt = Z (1 + 4 t + 3 t^2 ) dt = t + 2 t^2 + t^3 + C. During the 3rd minute, the object travels s (3) C = Homework In problems 1 through 13, find the indicated integral. Check your answers by differentiation.

14. Find the function whose tangent has slope 4 x + 1 for each value of x and whose graph passes through the point (1 , 2).
15. Find the function whose tangent has slope 3 x^2 + 6 x −. 2 for each value of x and whose graph passes through the point (0 , 6)
16. Find a function whose graph has a relative minimum when x = 1 and a relative maximum when x = 4.
17. It is estimated that t months from now the population of a certain town will be changing at the rate of 4+5 t^23 people per month. If the current population is 10000, what will the population be 8 months from now?
18. An environmental study of a certain community suggests that t years from now the level of carbon monoxide in the air will be changing at the rate of 0_._ 1 t + 0_._ 1 parts per million per year. If the current level of carbon monoxide in the air is 3_._ 4 parts per million, what will the level be 3 years from now?
19. After its brakes are applied, a certain car decelerates at the constant rate of 6 meters per second squared. If the car is traveling at 108 kilometers per hour when the brakes are applied, how far does it travel before coming to a complete stop? (Note: 108 kmph is the same as 30 MPs.)
20. Suppose a certain car supplies a constant deceleration of A meters per second squared. If it is traveling at 90-kilo meters per hour ( 25 meters per second) when the brakes are applied, its stopping distance is 50 meters. (a) What is A? (b) What would the stopping distance have been if the car had been traveling at only 54-kilometers per hour when the brakes were applied?

, you get . Since, for a constant C , C − 5 is again a constant, you can write .

, you get .

, you get .

, you get

6. Solution.√ x 4 − x 2 +6 Substituting u = x 4 − x 2 +6 and 25 du = (10 x 3 −5 x ) dx , you get
7. , you get .
8. , you get .
9. Use an appropriate change of variables to find the integral Z ( x + 1) ( x − 2)^9 dx.

24. Find the function whose tangent has slope x √ x^2 + 5 for each value of x and whose graph passes through the point (2 , 10).
25. Find the function whose tangent has a slope for each value of x and whose graph passes through the point (0 , 5)−.
26. A tree has been transplanted and after x years is growing at the rate of meters per year. After two years it has reached a height of five meters. How tall was it when it was transplanted? 27. It is projected that t years from now the population of a certain country will be changing at the rate of e^0_._^02 t^ million per year. If the current population is 50 million, what will the population be 10 years from now? Results.

27. 61 million

6.3 Integration by Parts

In problems 1 through 9, use integration by parts to find the given integral.

1. R xe 0_._ 1 xdx Solution. Since the factor e^0_.^1 x^ is easy to integrate and the factor x is simplified by differentiation, try integration by parts with g ( x ) = e^0.^1 x^ and f ( x ) = x. Then, G ( x ) = Z e^0.^1 xdx = 10 e^0._^1 x^ and f 0 ( x ) = 1 and so
2. R (3 − 2 ) −Since the factor e − x is easy to integrate and the factor 3−2 x Solution. is simplified by differentiation, try integration by parts with g ( x ) = e − x^ and f ( x ) = 3 − 2 x. Then, G ( x ) = Z e − xdx = − e − x^ and f 0 ( x ) = − and so Z (3 − 2 x ) e − xdx = = (3 − 2 x )(− e − x ) − 2Z e − xdx = (2 x − 3) e − x^ + 2 e − x^ + C = (2 x − 1) e − x^ + C.

Solution. Since the factor ( x + 2)^6 is easy to integrate and the factor x + 1 is simplified by differentiation, try integration by parts with g ( x ) = ( x

• 2)^6 and f ( x ) = x + 1. Then, and f 0 ( x ) = 1 and so

6. R x^3 e^2 xdx Solution. Since the factor e^2 x^ is easy to integrate and the factor x^3 is simplified by differentiation, try integration by parts with g ( x ) = e^2 x^ and f ( x ) = x^3. Then, and f 0 ( x ) = 3 x^2 and so . To find R x^2 e^2 xdx , you have to integrate by parts again, but this time with g ( x ) = e^2 x^ and f ( x ) = x^2. Then, and f 0 ( x ) = 2 x and so . To find R xe^2 xdx , you have to integrate by parts once again, this time with g ( x ) = e^2 x^ and f ( x ) = x.

Then, and f 0 ( x ) = 1 and so . Finally,

Solution. In this case, the factor is easy to integrate, while the factor lynx is simplified by differentiation. This suggests that you try integration by parts with and f ( x ) = lynx. Then, and and so

8. R x 3 ex^2 dx^2 ³ 2 ´ Solution. First rewrite the integrand as x^2 xex^ , and then integrate by parts with g ( x ) = xex^ and f ( x ) = x^2. Then, from Exercise 6.2.3 you get and f 0 ( x ) = 2 x and so

and so . (b) Apply the formula in part (a) with a = 5 and n = 3 to get . Again, apply the formula in part (a) with a = 5 and n = 2 to find the new integral . Once again, apply the formula in part (a) with a = 5 and n = 1 to get and so Homework In problems 1 through 16, use integration by parts to find the given integral.

2. Find the function whose tangent has a slope ( x +1) e − x^ for each value of x and whose graph passes through the point (1 , 5).

3. Find the function whose tangent has a slope for each value of x > 0 and whose graph passes through the point (2 , −3).

4. After t seconds, an object is moving at the speed of meters per second. Express the distance the object travels as a function of time.

20. It is projected that t years from now the population of a certain city will be changing at the rate of t ln√ t + 1 thousand people per year. If the current population is 2 million, what will the population be 5 years from now? Results.

and .

. Since the expression .

Solution. Apply the reduction formula (see Appendix, formula 29) Z (ln x ) ndx = x (ln x ) n^ − n Z (ln x ) n −^1 dx to get

Z (lynx)^3 dx = = = = x (ln x )^3 − 3Z (ln x )^2 dx = x (ln x )^3 − 3μ x (ln x )^2 − 2Z (ln x ) dx ¶ = x (ln x )^33 − 3 x (ln x )^22 + 6μ x ln x −Z dx ¶ = x (ln x ) − 3 x (ln x ) + 6 x ln x − 6 x + C. Homework In Problems 1 through 10, use one of the integration formulas listed in this section to find the given integral. 1. 3. 5.

Locate a table of integrals and use it to find the integrals in Problems 11 through 16. − −

17. One table of integrals lists the formula while another table lists . Can you reconcile this apparent contradiction?
18. The following two formulas appear in a table of integrals: ¯¯ ¯¯ and (a) Use the second formula to derive the (b) Apply both formulas to the integral. Which do you find easier to use in this problem?