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
.
To know more about Area bounded by the curve click here
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Answer:
2/3
Step-by-step explanation:
(6 -4) / (2 - (-1)) =(2) / (2+1) = 2/3
Answer:
4
Step-by-step explanation:
To find f(-1)g(-1) which is f(-1)*g(-1), first, we must find what are values of f(-1) and g(-1) first which can be found by substituting x = -1 in both functions.

Now we know that f(-1) is 2, next, we find g(-1).

Now we know that g(-1) is 2 too. Since the question wants to find f(-1)g(-1), therefore 2*2 = 4.
Hence, the answer is 4.
Answer:
8
Step-by-step explanation:
To find the mean, or average of a data set, add all the values together, then divide by the number of values.
The data set is: 6,7,9,9,9
1. Add all the values
Add 6,7,9,9 and 9
6+7+9+9+9
40
2. Divide by the number of values
Count how many numbers are in the data set. In this case, there are 5 numbers.
Divide 40 by 5.
40/5
8
The mean of the data set is 8.
I assume the equation described is:
( x + 6 ) / ( x^2 - 64 )
You can compare the degree of the numerator and denominator in a function that takes the form of this type of rational equation.
Here are the three rules
#1 (Correct Answer): When the degree of the numerator is smaller then the denominator the horizontal asymptote is y = 0
#2 If the degree of the numerator and denominator is the same, then you take the leading coefficient of the numerator (n) and denominator (d) to create the answer y = n / d in this equations case it would be 1 / 1 since variables technically have an invisible 1 in front of them since anything multiplied by 1 is its self, 1x = x
#3 When the degree of the numerator is greater then the degree of the denominator then this means that it does not have a horizontal asymptote.
Again the final answer is that the horizontal asymptote is y = 0