Explanation
The shaded area represents a segment. This can be solved with the formula below;
Since the triangle is an equilateral triangle, it implies that the angle subtended at the centre is 60 degrees. Also, the given radius is 7 cm
![\begin{gathered} =7^2(\frac{60}{360}\times3.14-\frac{1}{2}\times\sin 60)^{}_{} \\ =49(\frac{3.14}{6}-\frac{\sqrt[]{3}}{4}) \\ =4.43\operatorname{cm}^2 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%3D7%5E2%28%5Cfrac%7B60%7D%7B360%7D%5Ctimes3.14-%5Cfrac%7B1%7D%7B2%7D%5Ctimes%5Csin%2060%29%5E%7B%7D_%7B%7D%20%5C%5C%20%3D49%28%5Cfrac%7B3.14%7D%7B6%7D-%5Cfrac%7B%5Csqrt%5B%5D%7B3%7D%7D%7B4%7D%29%20%5C%5C%20%3D4.43%5Coperatorname%7Bcm%7D%5E2%20%5Cend%7Bgathered%7D)
Answer:
I think you have to first separate the integral:1/(1+v^2) + v/(1+v^2),
so the integral of the first term is ArcTan (v) and for the integral of the second term i recommend you to do a change of variable:
y= 1+v^2
so
dy= 2v
and
v= dy/2and then you substitute:v/(1+v^2) = (1/2)(dy/y)
and the integral is
(1/2) (In y)finally you plug in the initial variables:
(1/2)(In [1+v^2])
so the total integral is:
ArcTan (y) + (1/2)(In [1+v^2])
An algebraic expression doesnt have an equals sign or an inequality sign.
So,
A is an expression
B is an inequality
C is an equation
D is an in inequality
So the answer is A
