(2x^2 + x) / (x^2 + x - 2) dx Integrate by parts
(2x^2 + x) / (x^2 + x - 2) dx Integrate by parts
1) f(x)=6x-11/8x+1 2) f(x)=(5x-2)(2x+3)/x-4 3) f(x))=√¯8-5x
What is the smallest value of a if the area under the curve y=e^(ax) from -infinity to 4 is 11? I cannot seem to arrive at the answer using integrals. Please help.
∫ln(x) sin-1x dx
0∫ ∏ [sin (n + (1/2) ) X] / sin X dx equals : (2n +1 ) x ∏ / 2 0 ∏ n∏
using integration, find the area of each of the enclosed regions I've completed one correctly, but have no clue how to solve the other two can someone help? or set up the other two...
Using integration find the area of each of the enclosed regions how do I solve these?????? 1) x-axis and f(x)=2x-x^2 2) f(x)=x^2 and g(x)=x 3) f(x)=x^3-3x...
Using integration find the area of each of the enclosed regions how do I solve these?????? 1) x-axis and f(x)=2x-x^2 2) f(x)=x^2 and g(x)=x 3) f(x)=x^3-3x...
1) x-axis and f(x)=2x-x^2 2) f(x)=x^2 and g(x)=x 3) f(x)=x^3-3x and g(x)=x
find the area between y=0 and y=3(x3-x)
Find the average value of f on the given interval. f(x)=x^2 for 0≤x≤6
Find the average temperature Find the minimum temperature Find the maximum temperature
injected with this anti- inflammatory agent is given approximately by C(t)=42.54e^-.01042t, where t is the number of hours after the injection and 0≤t≤90. Question. What is the average...
What is the area of the region bounded by the curves y=2(x^2)-7 and y=2/x between x=1 and x=3? Please and thank you!
I need to solve the indefinite integral, however it must be by the u - sub method. ∫(x2+2)(x-1)7dx Im guessing this is not the best way, however by tutor tells...
integration ∫ √1 + sinx/4
Hello, can you please someone provide me solution for volume of revolution?: task: volume of revolution y=x^3, y=0, x=2 revolved about the x-axis I...
Intégral 3 d x / sqrt (4 - x^2) I think answer is 1.5 * pi / 6 + C Teacher says 3 * pi / 6 + C He divided each term under radical by 2 but I divided numerator by 2 also....
Suppose the marginal cost function (=supply curve) for these cell phone subscriptions is given by: p= C'(q)= 12.417 + 0.8884q p= domain input q= output. a)...
When I substitute u=x^2-5x+25 I get du=2x-5 dx. That gets me close to x dx. I would normally just pull out 1/2, but I'm confused how to deal with the -5. it looks like I should get a ln...