How to calculate calculus How much does ∫ (1,0) ∫ (2,0) 1 / 2xdxdy equal and how to calculate it? I'm learning how to calculate it

How to calculate calculus How much does ∫ (1,0) ∫ (2,0) 1 / 2xdxdy equal and how to calculate it? I'm learning how to calculate it

Multiple integral is to calculate one by one, first calculate inside
∫(1-0)∫(2-0) 1/2xdxdy=∫(2-0) [1/4*x^2](1-0) dy=∫(2-0) -1/4dy=[-1/4y](2-0)=1/2

How to calculate calculus
Since I haven't used calculus for nearly 10 years, I have forgotten all the knowledge of calculus. Now I need to use calculus to calculate. For example, if the sign of integral can't be typed out, let's use dictation. Assuming that the lower limit of integral is 0 and the upper limit is 5, then (x + 1) DX =? Please write out the calculation steps in detail,

∫(x+1)dx=∫xdx+∫1dx=(1/2)x²+x=(1/2)25+5=17.5

Ask for help with calculus
Integral (y ^ 2) * [e ^ (- x * y ^ 2)] dy
Offline x online x ^ 2

((-1/(4x))-x/2)e^(-x^5)+(1/(4x)+1/2e^(-x^3)
Ideas: using the step-by-step integration method;
Split (y ^ 2) * [e ^ (- x * y ^ 2)] into Y / (- 2x) and (- 2x) Ye ^ (- XY ^ 2)
So x can be regarded as a constant
Finally, the solution is given

Calculus calculation
0) 2 π y √ [1 + (y ') ^ 2] DX (surface area of the ball), the answer is 4 π R ^ 2, I want the process,

If y is a function of X, y ^ 2 + x ^ 2 = R ^, this is the relationship. When y is more than 0 and less than zero, then the result of calculation greater than 0 multiplied by 2 is the final result analysis: if we use the first type of surface fraction to calculate the spherical surface

How to calculate ∫ x ^ 2 √ 1 + x ^ 2DX integral?

As a result, we will be able to hold and and and and we will be able to be the last (and we will be one of the most most important and most of the first (1 + X \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\#8309; θ D θ - ∫

9.6 finding indefinite integral ∫ [cos (x / 2)] ^ 2DX

∫ [cos(x/2)]^2dx
=1/2*∫[2(cosx/2)^2-1+1 ]dx
=1/2*∫cosxdx+1/2*∫dx
=1/2*sinx+x/2+C

How to solve ∫ cos (x-1) DX, ∫ x ^ 3E ^ x ^ 2DX

∫ cos (x - 1) DX = ∫ cos (x - 1) d (x - 1) = sin (x - 1) + C ∫ X & # 179; e ^ (X & # 178;) DX -- > = ∫ UXe ^ u · Du / (2x) = (1 / 2) ∫ UE ^ u Du = (1 / 2) ∫ u de ^ u = (1 / 2) ue ^ u - (1 / 2) ∫ e ^ u Du = (1 / 2)

∫ x ^ 2DX ^ 2, how to calculate?

Let u = x ^ 2
Then ∫ x ^ 2DX ^ 2 = ∫ UDU = u ^ 2 / 2 = x ^ 4 / 2

∫(cos√x)^2dx

Let t = root x, x = T ^ 2, DX = 2tdt
The original formula = ∫ (cost) ^ 2 * 2tdt = ∫ (cos2t + 1) / 2 * 2tdt
=∫tcos2tdt+∫tdt
=1/2t*sin2t-1/2∫sin2tdt+t^2
=t/2*sin2t+1/4cos2t+t^2+C
Finally, change t to root X
I can't remember the formula. I don't know if it's right. Please correct it

Calculus calculation
Find the area of the figure enclosed by the curve y = SiNx, y = cosx and the straight line x = 0, x = - Pie / 4
Radical 2

It can be seen from the picture
The figure range is
-pi/4