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# 16 Calculus Problems (Math Problem Sample)

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There were around 16 calculus problems which i had to do while also explaining the step by step explanation. It was a univeristy level order and required in around 6 hours when i accepted it. I did it due time and also earned 5 stars from the client
Question as well as their explanation is written and well explained in the paper attached. source..

Content:

Student’s Name
Professor’s Name
Course Number
Date
Calculus Problems
Q4
xsec2x dx
Applying integration by parts:
Let u=x and dv=sec2(x)dx.
Then du=1dx
v=∫sec2(x)dx=tan(x)
∫xsec2xdx=x⋅tanx-tan∫x⋅1dx
=(xtan(x)-∫tan(x)dx)
Rewrite the tangent as tan(x)=sinxcosx
xtanx-∫tanxdx=xtanx-∫sinxcosx dx
As the derivative of cos(x) is present on the numerator, we can solve the integral:
=xtanx+∫-sinxcosx dx
=xtanx+lncosx+C
Q6
sinx ln(cosx) dx
Using substitution,
Let u=cos(x).
Then du=cosx'd=-sinxdx
sinx dx=-du.
The integral becomes
∫ln(cos(x))sin(x)dx=∫(-ln(u))du
=(-∫ln(u)du)
Use integration by parts:
-ln∫udu=-lnu⋅u-∫u⋅1udu
=-(uln(u)-∫1du)
=-u ln(u)-u
Replace u = cos(x)
∫lncosxsinxdx=-lncosxcosx+cosx+C
Or
∫lncosxsinxdx=1-lncosxcos(x)+C
Q9
ecosπxsinπx dx
By substitution,
Let u=cos(πx)
Then du=cosπx'd=-πsinπxdx
sin(πx)dx=-duπ
Therefore,
∫ecosπxsin(πx)dx=∫-euπdu
=(-∫euduπ)
The integral of the exponential function is ∫eudu=eu:
-∫euduπ=-euπ
Recall that u=cos(πx):
-euπ=-ecosπxπ
Therefore,
∫ecosπxsinπxdx=-ecosπxπ+C
Q10
Using the trapezoidal rule for 3 subintervals,
A=18+(-6)27-0+-3-6216-7+20-3225-16
A=42-40.5+76.5
A=78
Q15
23x1+23x dx
Rewriting,
∫8x8x+1 dx
∫8x8x+1dx=8 ∫x8x+1dx
Rewrite the numerator of the integrand as
x=188x+1-18
and split the fraction
8∫x8x+1dx=818-188x+1dx
Integrate term by term:
818-188x+1dx=8x8-8 188x+1dx
=x-1888x+1
=x-ln8x+18+C
Homework 6.7
Q3
x233x3+7 dx
Let u=3x3+7
Then
du=9x2dx
x2dx=du9
Thus,
x233x3+7 dx=3u9 du
Or,
=193udu=19u13du
Apply the power rule
=3u4349=u4312
Recall that u=3x3+7
=3x3+74312+C
Homework 6.4
Q5
02fxdx+25fxdx
Single integral:
=05fxdx
Q8
51fxdx+85fxdx
Single integral:
=-15fxdx-58fxdx
=-15fxdx+58fxdx
=-18fxdx
Q7
limn→∞3nk=1n2+3kn4
Here,
Δx⟷3n is the rectangle width.
f(a+kΔx)⟷2+k⋅3n
So the left endpoint a is 2.
And fx=x
we can solve for the right endpoint b because we know that Δx=b-an
3n=b-2n
b=5
So, the integral will be:
=25x4dx
And calculating by calculator, we get:
25x4dx=618.6
Q13
gx=2sinx over [0, π]
As,
abgxdx=limn→∞i=1ng(a+i∆x)∆x
Given:
gx=2sinx
Interval:
[0, π]
Δx=b-an=π-0n=πn
So, the area is computed using a definite integral and a definite integral is calculated using the limit of a Riemann sum.
Area=abgxdx=limn→∞i=1ng(a+i∆x)∆x
=limn→∞i=1ng(0+iπn)πn
=limn→∞i=1nπn giπn
=limn→∞i=1nπn 2siniπn
Q15
f'x=1x ;f0=3
y...

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