Obtain the Fourier series of f(x) = x2 in the interval (0, 2π)?
Answers
Answer:
Answer:
Correct Question -
The circumference of two circle are in the ratio 2 : 3. Find the ratio of their areas.
Given -
Ratio of their circumference = 2:3
To find -
Ratio of their areas.
Formula used -
Circumference of circle
Area of circle.
Solution -
In the question, we are provided, with the ratios of the circumference of 2 circles, and we need to find the ratio of area of those circle, for that first we will use the formula of circumference of a circle, then we will use the formula of area of circles. We will be writing 1 equation in it too.
So -
Let the circumference of 2 circles be c1 and c2
According to question -
c1 : c2
Circumference of circle = 2πr
where -
π = \tt\dfrac{22}{7}
7
22
r = radius
On substituting the values -
c1 : c2 = 2 : 3
2πr1 : 2πr2 = 2 : 3
\tt\dfrac{2\pi\:r\:1}{2\pi\:r\:2}
2πr2
2πr1
= \tt\dfrac{2}{3}
3
2
\tt\dfrac{r1}{r2}
r2
r1
= \tt\dfrac{2}{3}
3
2
\longrightarrow⟶ [Equation 1]
Now -
Let the areas of both the circles be A1 and A2
Area of circle = πr²
So -
Area of both circles = πr1² : πr2²
On substituting the values -
A1 : A2 = πr1² : πr2²
\tt\dfrac{A1}{A2}
A2
A1
= \tt\dfrac{(\pi\:r1)}{(\pi\:r2)}^{2}
(πr2)
(πr1)
2
\tt\dfrac{A1}{A2}
A2
A1
= \tt\dfrac{(r1)}{(r2)}^{2}
(r2)
(r1)
2
\tt\dfrac{A1}{A2}
A2
A1
= \tt\dfrac{(2)}{(3)}^{2}
(3)
(2)
2
[From equation 1]
So -
\tt\dfrac{A1}{A2}
A2
A1
= \tt\dfrac{4}{9}
9
4
\therefore∴ The ratio of their areas is 4 : 9
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Answer:
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