If the area of the region bounded by the curve
and the line
is
Sq units, then the value of
will be
.
<h3>What is area of the region bounded by the curve ?</h3>
An area bounded by two curves is the area under the smaller curve subtracted from the area under the larger curve. This will get you the difference, or the area between the two curves.
Area bounded by the curve
We have,
⇒ 
,
Area of the region
Sq units
Now comparing both given equation to get the intersection between points;

So,
Area bounded by the curve
![\frac{256}{3} =\[ \int_{0}^{4a} \sqrt{4ax} \,dx \]](https://tex.z-dn.net/?f=%5Cfrac%7B256%7D%7B3%7D%20%3D%5C%5B%20%20%5Cint_%7B0%7D%5E%7B4a%7D%20%5Csqrt%7B4ax%7D%20%20%5C%2Cdx%20%5C%5D)
![\frac{256}{3}= \[\sqrt{4a} \int_{0}^{4a} \sqrt{x} \,dx \]](https://tex.z-dn.net/?f=%5Cfrac%7B256%7D%7B3%7D%3D%20%20%20%5C%5B%5Csqrt%7B4a%7D%20%20%5Cint_%7B0%7D%5E%7B4a%7D%20%5Csqrt%7Bx%7D%20%20%5C%2Cdx%20%5C%5D)
![\frac{256}{3}= 2\sqrt{a} \left[\begin{array}{ccc}\frac{(x)^{\frac{1}{2}+1 } }{\frac{1}{2}+1 }\end{array}\right] _{0}^{4a}](https://tex.z-dn.net/?f=%5Cfrac%7B256%7D%7B3%7D%3D%202%5Csqrt%7Ba%7D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5Cfrac%7B%28x%29%5E%7B%5Cfrac%7B1%7D%7B2%7D%2B1%20%7D%20%7D%7B%5Cfrac%7B1%7D%7B2%7D%2B1%20%7D%5Cend%7Barray%7D%5Cright%5D%20_%7B0%7D%5E%7B4a%7D)
![\frac{256}{3}= 2\sqrt{a} \left[\begin{array}{ccc}\frac{(x)^{\frac{3}{2} } }{\frac{3}{2} }\end{array}\right] _{0}^{4a}](https://tex.z-dn.net/?f=%5Cfrac%7B256%7D%7B3%7D%3D%202%5Csqrt%7Ba%7D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5Cfrac%7B%28x%29%5E%7B%5Cfrac%7B3%7D%7B2%7D%20%7D%20%7D%7B%5Cfrac%7B3%7D%7B2%7D%20%7D%5Cend%7Barray%7D%5Cright%5D%20_%7B0%7D%5E%7B4a%7D)
![\frac{256}{3}= 2\sqrt{a} *\frac{2}{3} \left[\begin{array}{ccc}(x)^{\frac{3}{2}\end{array}\right] _{0}^{4a}](https://tex.z-dn.net/?f=%5Cfrac%7B256%7D%7B3%7D%3D%202%5Csqrt%7Ba%7D%20%2A%5Cfrac%7B2%7D%7B3%7D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%28x%29%5E%7B%5Cfrac%7B3%7D%7B2%7D%5Cend%7Barray%7D%5Cright%5D%20_%7B0%7D%5E%7B4a%7D)
On applying the limits we get;
![\frac{256}{3}= \frac{4}{3} \sqrt{a} \left[\begin{array}{ccc}(4a)^{\frac{3}{2} \end{array}\right]](https://tex.z-dn.net/?f=%5Cfrac%7B256%7D%7B3%7D%3D%20%5Cfrac%7B4%7D%7B3%7D%20%5Csqrt%7Ba%7D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%284a%29%5E%7B%5Cfrac%7B3%7D%7B2%7D%20%20%5Cend%7Barray%7D%5Cright%5D)



⇒ 

Hence, we can say that if the area of the region bounded by the curve
and the line
is
Sq units, then the value of
will be
.
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344 thousands times 1/10=34,400
Answer:
YES! we conclude that f(x) = 1/3x + 5 and g(x) = 3x - 15 are inverse functions.
Step-by-step explanation:
Given
Given that the function f(x) and g(x) are inverse functions.


To determine
Let us determine whether f(x) = 1/3x + 5 and g(x) = 3x - 15 are inverse functions.
<u>Determining the inverse function of f(x) </u>
A function g is the inverse function of f if for y = f(x), x = g(y)

Replace x with y

Solve for y

Therefore,
YES! we conclude that f(x) = 1/3x + 5 and g(x) = 3x - 15 are inverse functions.