Math Problem: Integration?

Question

need a solution to this:

Root of (a2-x2 )dx

Answer 1

I'm sure someone else will do better than I will on this, but if it helps, according to the table of integrals in my calc textbook, it should be:

x²/2 √(a² - x²) + a²/2 sin-¹(x/a) + C

That's sin^-1, aka arcsin.

I'm way past the point where I could explain how to derive it without reviewing unfortunately.

Answer 2

First discover the place the two curves intersect with the aid of fixing x^2 = 3 -2x (x = -3, one million) Now the area often is the fundamental of three- 2x - x^2 over (x = -3, one million) 3x - x^2 - one million/3 x^3 over (x = -3, one million) 3 - one million -one million/3 + 9 - 9 +9 = 10 2/3

Answer 3

Root of(a^2-x^2)dx=roof of a^2dx-roof of x^2dx
'' =(a^n+1)/n+1-(x^n+1)/n+1

or n=2
Roof of(a^2-x^2)dx=(a^2+1)/2+1-(x^2+1)/n+1
'' =a^3/3-x^3/3

Roof of (a^2-x^2) =1/3(a^3-x^3)+K(constant)

Answer 4

Let x=a*sin(t), then dx=a*cos(t)*dt; thus y=sqrt(a^2-x^2)*dx = a^2*cost*cost*dt
cost*cost=(1+cos(2t))/2, as cos(2t)=cost*cost-sint*sint;
thus y=0.5*a^2*(1+cos2t)*dt; I=0.5*a^2*(C+t+0.5sin2t) – integration done!

Answer 5

ok.substitute x=asin(y) and you will get
int. sqrt(a^2-x^2) dx
= int. sqrt[a^2-a^2{sin(y)}^2] a cos(y) dy
= a^2 [ int. sqrt[1-{sin(y)}^2 cos(y) dy]
= a^2 [ int. sqrt[cos(y)]^2 cos(y) dy]
= a^2 [ int. cos(y)^2 dy]
= a^2/2 [int. 1+ cos(2y) dy]
= a^2/2 [y+ sin(2y)/2 ] + C , where C is a constant.
But we know that y=arcsin(x/a),and
sin(2y)=2sin(y)cos(y)=2sin(arcsin(x/a))cos(arcsin(x/a))
=2(x/a)(sqrt(a^2-x^2)/a)
Hence, int. sqrt(a^2-x^2) dx
= x[sqrt(a^2-x^2)]/2 + a^2(arcsin(x/a))/2 + C #

Answer 6

let u=rt(a^2-x^2) and dv=dx
du=-x/rt(a^2-x^2)dx and v=x
integrating by parts
I=rt(a^2-x^2)*x-intx(-x)dx
/rt(a^2-x^2)
=xrt(a^2-x^2-int(-x^2)dx
/rt(a^2-x^2)
xrt(a^2-x^2)-int(a^2-x^2-a^2)dx
/rt(a^2-x^2)
=xrt(a^2-x^2)-int(a^2-x^2)dx
/rt(a^2-x^2)+a^2indx/rt(a^2-x^2)
I=xrt(a^2-x^2)-I+a^2sin^-1(x/a)+k
2I=xrt(a^2-x^2)+a^2sin^-1(x/a)+k
I=(x/2)rt(a^2-x^2)+(a^2/2)sin^-1(x/a)+C