1 and 5 we havePROBLEMS 14.7Section 14.7 The Fundamental Theorem of integral Caloukus9992.(9)he rate of5x dx43.where f(r)-3.4.uffices to-5x dx(2r - 3) dx44. Evaluate6.xpx5.o² +7.+ 4y + 4) dy45. Suppose f(x)D-1 dr. Evaluate3 dt.8.&p (66-)(2t-3r2) dt46. Evaluate|9.10.+ xpxp47. If3/212.f(x) dx 5 and3.x23/211.%3DVx4 dx8.13.14.48. Iff) dx =5, and*9 = xp (x)fx) dx = 2, findxp11/216.xp (+x+)costxpxp (x)f15.49. Evaluate(z + 1)* dzxp (- x^)|181 18.xp(Hint: It is not necessary to find 217.xp xpence-22x²(x³ – 1) dxxp (/1-X - €/)19.20.(x+ 2)³ dxe -e1= (x)fIn discussing traffic safety, Shonles50. Suppose that f(x) =(30022.dt where x e. Find f'(x).%3Det +e-t4.Ap –51. Severity Indexconsiders how much acceleration a person can tolerate in a crashso that there is no major injury. The severity index is defined as21.2.xpBronbxpLonoiens23.24.CTX - 1xp5/212 dt= T'S26.| (3x2 + 4x)(x + 2x?)* dxwhere a (a Greek letter read "alpha") is considered a constantinvolved with a weighted average acceleration, and T is theduration of the crash. Find the severity index.25.xpbioqne3.(x+3)220/328.nou-1/327.xpV3x + 5 dx/10- 3p dpav9? +3 dgbp30.29.1/331.32.dx(x² + 1)2/3-xP I + pXLA* /9.33.34.(m + ny) dy52. StatisticsIn statistics, the mean u (a Greek letter reacxp“mu") of the continuous probability density function f definthe interval [a, b] is given by-35.36.8|x| dxxp-2xp (x) fx37.| 3(x-2 +x-3 –x-4) dxand the variance o? (o is a Greek letter read "sigma") is gxp39.(x+1)e+2* dx1.|38.= ,0xp (X) f-(ri – x) =Compute u and then o? if a = 0, b = 1, and f (x) = 6(xThe economist Paretoa.%3D%3D95xpIn e40.53. Distribution of Incomes2++6x* +x³ + 8x2 +x+ 5x3 + 5x + 1an empirical law of distribution of higher incomes thatnumber N of persons receiving x or more dollars. If41.xpNP=-Ar-Bxp42.dx (Hint: Multiply the integrand by )J. I. Shonle, Environmental Applications of General Physics CAddison-Wesley Publishing Company, Ic., 1975).G. Tintner, Methodology of Mathematical Economics and Ec(Chicago: University of Chicago Press, 1967), p. 16.

Question

1 and 5 we havePROBLEMS 14.7Section 14.7 The Fundamental Theorem of integral Caloukus9992.(9)he rate of5x dx43.where f(r)-3.4.uffices to-5x dx(2r - 3) dx44. Evaluate6.xpx5.o² +7.+ 4y + 4) dy45. Suppose f(x)D-1 dr. Evaluate3 dt.8.&p (66-)(2t-3r2) dt46. Evaluate|9.10.+ xpxp47. If3/212.f(x) dx 5 and3.x23/211.%3DVx4 dx8.13.14.48. Iff) dx =5, and*9 = xp (x)fx) dx = 2, findxp11/216.xp (+x+)costxpxp (x)f15.49. Evaluate(z + 1)* dzxp (- x^)|181 18.xp(Hint: It is not necessary to find 217.xp xpence-22x²(x³ – 1) dxxp (/1-X - €/)19.20.(x+ 2)³ dxe -e1= (x)fIn discussing traffic safety, Shonles50. Suppose that f(x) =(30022.dt where x > e. Find f'(x).%3Det +e-t4.Ap –51. Severity Indexconsiders how much acceleration a person can tolerate in a crashso that there is no major injury. The severity index is defined as21.2.xpBronbxpLonoiens23.24.CTX - 1xp5/212 dt= T'S26.| (3x2 + 4x)(x + 2x?)* dxwhere a (a Greek letter read &#34;alpha&#34;) is considered a constantinvolved with a weighted average acceleration, and T is theduration of the crash. Find the severity index.25.xpbioqne3.(x+3)220/328.nou-1/327.xpV3x + 5 dx/10- 3p dpav9? +3 dgbp30.29.1/331.32.dx(x² + 1)2/3-xP I + pXLA* /9.33.34.(m + ny) dy52. StatisticsIn statistics, the mean u (a Greek letter reacxp“mu&#34;) of the continuous probability density function f definthe interval [a, b] is given by-35.36.8|x| dxxp-2xp (x) fx37.| 3(x-2 +x-3 –x-4) dxand the variance o? (o is a Greek letter read &#34;sigma&#34;) is gxp39.(x+1)e+2* dx1.|38.= ,0xp (X) f-(ri – x) =Compute u and then o? if a = 0, b = 1, and f (x) = 6(xThe economist Paretoa.%3D%3D95xpIn e40.53. Distribution of Incomes2++6x* +x³ + 8x2 +x+ 5x3 + 5x + 1an empirical law of distribution of higher incomes thatnumber N of persons receiving x or more dollars. If41.xpNP=-Ar-Bxp42.dx (Hint: Multiply the integrand by )J. I. Shonle, Environmental Applications of General Physics CAddison-Wesley Publishing Company, Ic., 1975).G. Tintner, Methodology of Mathematical Economics and Ec(Chicago: University of Chicago Press, 1967), p. 16.

Accepted Answer

Given ∫035 dxwe know that ∫a.dx=a.xSo, ∫035 dx=5.[x]03
=5.[3-0]=5*3=15∴∫035 dx=15

In 5 dx 1. 5x dx 3. (2r - 3) dx 5. -3