4 pages/≈1100 words
Mathematics & Economics
16 Calculus Problems (Math Problem Sample)
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..
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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