Find the area of the region between the curve y^2=4x and the line x=3.

ok, sketch it.

area=INT2ydx from x=0 to 3

area=INT 2*sqrt(4x) dx
= INT 2*2sqrtx dx from x=0 to x^3

= 4*2/3 x^3 over limits
= 8/3 *27

check my work

I sorry but I don't undestand. What would be f(x) and g(x)?

I have no idea what you mean by F(x) or g(x).

I'm sorry. For the equation that I have to use for this it is:

Integral symbol(f(x)-g(x))dx

Sorry I posted into the wrong conversation, please ignore me. Thank you.

f(x)=sqrt(4x) g(x)=-sqrt(4x)

f(x)-g(x)=2*f(x)

Why is g(x)=-sqrt(4x)?

sketch it!

sqrt4x=+- 2sqrtx

Oh I forgot about the ± sign. THANK YOU!!

To find the area of the region between the curve y^2=4x and the line x=3, we need to first determine the points of intersection between the curve and the line.

The given curve is y^2 = 4x, which is a parabola opening towards the right. The given line is x = 3, which is a vertical line passing through x = 3 on the x-axis.

To find the points of intersection, we can substitute x = 3 into the equation of the curve:

y^2 = 4 * 3
y^2 = 12

Taking the square root of both sides, we have:

y = ± √12
y = ± 2√3

So, the curve intersects the line at two points: (3, 2√3) and (3, -2√3).

To find the area between the curve and the line, we need to integrate the difference between the curve and the line with respect to x from the x-coordinate of the left intersection point to the x-coordinate of the right intersection point.

Let's denote the left intersection point as A (3, 2√3) and the right intersection point as B (3, -2√3).

To determine the limits of integration, we need to find the x-coordinate of the intersection points. In this case, both points have an x-coordinate of 3.

Now, let's set up the integral for the area:

A = ∫ [B-A] y^2 dx

Where [B-A] denotes the limits of integration.

Substituting y^2 = 4x into the equation, we have:

A = ∫ 4x dx

Integrating with respect to x, we get:

A = 2x^2 + C

Evaluating the integral between the limits of integration (x = 3 to x = 3), we have:

A = [2(3)^2 + C] - [2(3)^2 + C]
A = 0

Therefore, the area of the region between the curve y^2 = 4x and the line x = 3 is 0.