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 ![= \[ \int_{0}^{4a} y \,dx \]](https://tex.z-dn.net/?f=%3D%20%20%20%5C%5B%20%5Cint_%7B0%7D%5E%7B4a%7D%20y%20%5C%2Cdx%20%5C%5D)
![\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}= \[2\sqrt{a} \int_{0}^{4a} x^{\frac{1}{2} } \,dx \]](https://tex.z-dn.net/?f=%5Cfrac%7B256%7D%7B3%7D%3D%20%20%20%5C%5B2%5Csqrt%7Ba%7D%20%20%5Cint_%7B0%7D%5E%7B4a%7D%20x%5E%7B%5Cfrac%7B1%7D%7B2%7D%20%7D%20%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 .
To know more about Area bounded by the curve click here
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31-62=152+2-6(3+21)
31-62=152+2-18+126
-31 =262
Answer:
269
Step-by-step explanation:
Solve 12x + 6 2 9x + 12.
A. X2
B. xs2
O
C. X26
0
D. Xs6
Answer:
600
Step-by-step explanation:
since it is a right triangle based pyramid, 2 of them will make a cube. so, it is bwh / 2.
base = 24
width = 10
height = 5
volume = 1200
volume of pyramid = 600
Answer:
Hello! answer: 252
Step-by-step explanation:
To find parelogram area you do base × height so... 21 × 12 = 252 therefore 252 is the area Hope that helps!
