Question

If the area of the region bounded by the curve y^2 =4ax and the line x= 4a is 256/3 Sq units, using integration find the value of a, where a > 0.

Answer 1

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

How to determine the value of a?

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

$A =2 \int\limits^{4a}_0 {\sqrt y} \, dx$

This gives

$A =2 \int\limits^{4a}_0 {\sqrt {4ax}} \, dx$

Remove the constant

$A = 2\sqrt{4a}\int\limits^{4a}_0 {\sqrt {x}} \, dx$

Integrate

$A = 2\sqrt{4a} * \frac 23 x^\frac32 |\limits^{4a}_0$

Evaluate the exponent

$A = 4\sqrt{a} * \frac 23 x^\frac32 |\limits^{4a}_0$

Expand

$A = 4\sqrt{a} * \frac 23 (4a)^\frac32$

Rewrite as:

$A = 4 * a^\frac12 * \frac 23 * 4^\frac32 * a^\frac32$

Evaluate the exponents by the laws of indices

$A = 4 * \frac 23 * 8 * a^2$

Recall that the area is:

A = 256/3

So, we have:

$\frac{256}3 = 4 * \frac 23 * 8 * a^2$

Evaluate the quotients

$256 = 4 * 2 * 8 * a^2$

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:

brainly.com/question/19053586

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