The value of a using integration is 2, such that the area of the region bounded by the curve y^2 =4ax and the line x= 4a is 256/3 Sq units
The given parameters are:
Curve, y² = 4ax
Line, x = 4a
Substitute x = 4a in y² = 4ax
y² = 4a * 4a
Take the square root of both sides
y = ±4a
This means that the point of intersection between the line and the curve is (4a, -4a) and (4a, 4a)
Remove the negative interval
So, we have: (4a, 0) and (4a, 4a)
The area is then calculated as
[tex]A =2 \int\limits^{4a}_0 {\sqrt y} \, dx[/tex]
This gives
[tex]A =2 \int\limits^{4a}_0 {\sqrt {4ax}} \, dx[/tex]
Remove the constant
[tex]A = 2\sqrt{4a}\int\limits^{4a}_0 {\sqrt {x}} \, dx[/tex]
Integrate
[tex]A = 2\sqrt{4a} * \frac 23 x^\frac32 |\limits^{4a}_0[/tex]
Evaluate the exponent
[tex]A = 4\sqrt{a} * \frac 23 x^\frac32 |\limits^{4a}_0[/tex]
Expand
[tex]A = 4\sqrt{a} * \frac 23 (4a)^\frac32[/tex]
Rewrite as:
[tex]A = 4 * a^\frac12 * \frac 23 * 4^\frac32 * a^\frac32[/tex]
Evaluate the exponents by the laws of indices
[tex]A = 4 * \frac 23 * 8 * a^2[/tex]
Recall that the area is:
A = 256/3
So, we have:
[tex]\frac{256}3 = 4 * \frac 23 * 8 * a^2[/tex]
Evaluate the quotients
[tex]256 = 4 * 2 * 8 * a^2[/tex]
Divide both sides by 64
4 = a²
Take the square root of both sides
2 = a
Rewrite as:
a = 2
Hence, the value of a is 2
Read more about integration at:
https://brainly.com/question/19053586
#SPJ1
