[FULL ACCESS] Integral Calculus and Advanced Math Solutions

This assignment provides step-by-step solutions to various integral calculus and advanced math problems. Students can access detailed explanations and calculations for topics such as integration by parts, exact value calculations, and more. The provided solutions cover a range of mathematical concepts, making it an essential resource for students looking to improve their understanding and skills in integral calculus and advanced math.

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Table of Contents
Question 1:.......................................................................................................................................3
Find the following equations.......................................................................................................3
Question 2:.......................................................................................................................................4
Differentiate................................................................................................................................4
Question 3: ......................................................................................................................................5
The energy E stored in a spring is given by the formula E = 200 x ^ 2 where x is the
compression. The rate of change of the energy with respect to the compression is called the
thrust T of the spring...................................................................................................................5
Question 4:.......................................................................................................................................6
Find dy/ dx at the point (0, 2) given that y ^ 3 = 8 + xe^ y ........................................................6
Question 5. ......................................................................................................................................6
The total force due to a distributed load acting on a beam from x = a to x = b is given by........6
∫a-b f(x)dx....................................................................................................................................6
Question 6:.......................................................................................................................................8
Evaluate the following integrals using integration by parts:.......................................................8

Question 1:
Find the following equations
(a) limx→−5 (25 – x^2 / x^2 + 2 x- 15)
Ans: => 25-x ^ 2 / x^2 + 2 x-15
= - x-5 / x-3
= 25- x^2 / x^2 + 2 x -15
Factors: -x^2 + 25 : (x+5)(x-5)
= - (x+5)(x-5) / x^2 + 2 x - 15
Factor x ^ 2 + 2 x – 15
=-(x+5)(x-5) / (x-3)(x+5)
cancel the common factor : x+5
= - x-5/ x-3
= lim x->-5 (-x-5 / x-3)
Putting value x= -5
= – -5-5 / -5-3
= - 10 / 8
cancel common factor:
= -5/4
(b) limx→∞ 4x^2 – 19 / 3x^2 + 11
Ans: limx→∞ (4 x^2-19 / 3 x^2+11)
Divide by highest denominator power:
4- 19/x^2 / 3+ 11/x^2
= limx→∞ (4 - 19/x^2 / 3+ 11/x^2)
= limx→∞ (4 - 19/x^2) / limx→∞ (3+ 11/x^2)
= limx→∞ (4 – 19/x^2) = 4 : simplify
= limx→∞ (3+ 11/x^2) = 3 : simplify
= 4 / 3

Question 2:
Differentiate
(a) e ^ sin(x 2 ) ;s
Ans : d / dx (e ^sin x^2)
= e ^sin (x^2) cos(x^2). 2X
Apply chain rules:
= d / du (e^u)d/dx(sin(x^2))
= d / du (e^u)= e^u
= e ^u
apply this common derivations:
d / d x (sin x^2) = cos (x^2).2x
= e ^u cos(x^2).2x
substitute back u= sin(x^2)
= e^sin(x^2) cos(x^2).2x
(b) ln sin−1 x 3 , where sin−1 x is an inverse function for sin x, i.e. sin−1 x = arcsin x
Ans: d/ dx (sin ^-1 x^3)
= d/dx( arcsin(x^3) = 3x^2 / √1-x^6
= d / dx (arcsin (x^3))
apply chain rule:
d/ du ( arcsin (u)) d/ dx(x^3)
= d/ dx (x^3) = 3x^2
= 1/ √1- u^2 . 3X^2
substitute back u = x^3
= 1/ √1- (x^3)^2 . 3X^2
Simplify :
= 1/ √1- (x^3)^2 . 3X^2 : 3x^2 / √1-x^6
= 3x^2 / √1-x^6

Question 3:
The energy E stored in a spring is given by the formula E = 200 x ^ 2 where x is the
compression. The rate of change of the energy with respect to the compression is
called the thrust T of the spring.
(a) Express the thrust T of the spring in terms of x;
Ans: R indirectly propositional to 1/E
T = c. E
T= thrust
C = Compression
E= energy
(b) Find the energy E at x = 1.5;
Ans: E= 200. (1.5) ^ 2
= 2.25 x 200
= 450 (energy)

Find the thrust T at x = 1.5;
Ans: T= 450 x 1.5 {thrust = Compression x energy}
= T= 675
Find the compression x when the energy E = 400;
Ans: E= 200(x)^2
= 400 = 200 X^2
x^2 = 400 / 200
x^ 2= 2
x= √ x
(e) Find the compression x when the thrust T = 100
Ans: Compression = Thrust / Energy
x= 100 / 450
x= 0.2
Question 4:
Find dy/ dx at the point (0, 2) given that y ^ 3 = 8 + xe^ y .
Ans: d / dx (y . 3) = d / dx (8+ xe^y)
= d/ dx (y . 3) = 3d/dx(y)
= d/dx(8+xe^y)= e^y+e^y x d/dx(y)
= 3 d/dx(y) = e^y+e^y x d/dx(y)
islolate in terms of y:
y' = e^y / -xe^y +3
d/dx(y)= e^y / -xe^y +3
Question 5.
The total force due to a distributed load acting on a beam from x = a to x = b is given by
∫a-b f(x)dx
where f(x) is the force at the point x.
Ans : ∫a->b f(x)dx
= ∫a->bfxdx
Compuet the integral : ∫ fx dx= fx^2/2 +c
= f.∫ xdx

= f x^1+1/ 1+1
Simplify:
=fx^2/2
(a) Find the indefinite integral of the force
f(x) = 6x − 2x 3 − x √ x + 3e ^x
i.e. ∫ f(x)dx
Ans: ∫ ( 6 x − 2x 3 − x √ x + 3e ^x)
= apply the sum rules:
= ∫ 6x dx- ∫ 2x^3dx - ∫ x √ x dx + ∫ 3e ^x dx
= ∫ 6x dx = 3x^2
= ∫ 2x^3 dx= x^4 / 2
=∫ x √ x dx = ½ x^2 √ x – 1/10 x^5/2
=∫ 3e^x dx= 3e^x
Simplify:
= 3x^2-x^4/2-1/2x^2 √ x – 1/10 x^5/2 + 3e^x
(b) Hence calculate the exact value of the total force if a = 0 and b = 1, i.e. evaluate the
integral
F =∫ 1 - 0 (6x − 2x ^3 − x √ x + 3e ^ x ) dx
Ans: ∫ 1 – 0 (6x − 2x ^3 − x √ x + 3e ^ x ) dx
Compute integral:
∫ 1 – 0 (6x − 2x ^3 − x √ x + 3e ^ x ) dx = 3x^2-x^4 / 2-1 / 2x^2 √ x +1/10 x^5/2 + 3e^x+c
Apply chain rules:
= ∫6xdx=3x^2
= ∫2x^3dx={x^4}/{2}
= ∫2x^3dx
taking constant out:
= 2 ∫ x^3dx
simplify by using power rule:
= 2. {x^{3+1} /{3+1}: {x^4}/ {2}
= {x^4} / {2}
= ∫x √ x dx={1}/{2}x^2{x}-{1}/ {10}x^{5}/ {2}}

Compute and simplify:
=3e+ {21}/ {10}-3
= =3e-[9} / {10}
Question 6:
Evaluate the following integrals using integration by parts:
(a) ∫ in x / x^5 dx;
Ans: ∫ x / x^5 dx = -{1} / {3x^3}+C
= ∫ x / x^5 dx
= x / x^5= x^{-4}
= ∫ x^{-4} dx
applying power rule:
= x^-4+1 / -4 +1
Simplify:
x^-4+1 / -4 +1= -1/3x^3
= -1/3x^3
= -1/3x^3 +c
(b) ∫ (2x + 5) cos 4x dx.
Ans: Apply integration u= (2x+5), v= cos(4x)
= {1} / {4} sin (4x)(2x+5)- ∫ ½ sin(4x) dx
= 1/ 2 sin (4x) (2x+5) - (-1/8 cos(4x))
Simplify:
= ¼ sin (4x) (2x+5)+ 1/8cos (4x)
= ¼ sin (4x) (2x+5) + 1/8 cos (4x) +c