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4 pages/≈1100 words
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MLA
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Mathematics & Economics
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Math Problem
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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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