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The area bounded by the curve y2 = 9x and the lines x = 1, x = 4 and y = 0 in the first quadrant is
- a)7
- b)14
- c)28
- d)14/3
Correct answer is option 'B'. Can you explain this answer?
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The area bounded by the curve y2 = 9x and the lines x = 1, x = 4 and y...
**Problem:**
The area bounded by the curve $y^2 = 9x$ and the lines $x = 1$, $x = 4$ and $y = 0$ in the first quadrant is
A) 7
B) 14
C) 28
D) 14/3
**Solution:**
The given curve is a parabola of the form $y^2 = 4ax$ which opens towards the right. Here, $a = \dfrac{9}{4}$.
The given lines are $x=1$, $x=4$ and $y=0$.
The points of intersection of the given curve with the given lines can be found by substituting $x=1$ and $x=4$ in the equation of the curve. We get the points of intersection as $(1,3)$ and $(4,6)$.
Hence, the required area is given by
$$\int_1^4 \dfrac{y^2}{9} dx = \dfrac{1}{9} \int_1^4 9x dx = \dfrac{1}{9} \left[\dfrac{9x^2}{2}\right]_1^4 = \dfrac{1}{2} (4^2 - 1^2) = \boxed{14}$$
Hence, the correct option is (B).
**Note:**
The area bounded by the curve $y^2 = 4ax$ and the lines $x = a$, $x = b$ and $y = 0$ in the first quadrant is given by
$$\dfrac{1}{2} (b-a) \times 2a = a(b-a)$$
Here, $a = \dfrac{9}{4}$ and $b = 4$. Hence, the required area is $a(b-a) = \dfrac{9}{4} \times \dfrac{7}{4} = \dfrac{63}{16}$.
The area bounded by the curve $y^2 = 9x$ and the lines $x = 1$, $x = 4$ and $y = 0$ in the first quadrant is
A) 7
B) 14
C) 28
D) 14/3
**Solution:**
The given curve is a parabola of the form $y^2 = 4ax$ which opens towards the right. Here, $a = \dfrac{9}{4}$.
The given lines are $x=1$, $x=4$ and $y=0$.
The points of intersection of the given curve with the given lines can be found by substituting $x=1$ and $x=4$ in the equation of the curve. We get the points of intersection as $(1,3)$ and $(4,6)$.
Hence, the required area is given by
$$\int_1^4 \dfrac{y^2}{9} dx = \dfrac{1}{9} \int_1^4 9x dx = \dfrac{1}{9} \left[\dfrac{9x^2}{2}\right]_1^4 = \dfrac{1}{2} (4^2 - 1^2) = \boxed{14}$$
Hence, the correct option is (B).
**Note:**
The area bounded by the curve $y^2 = 4ax$ and the lines $x = a$, $x = b$ and $y = 0$ in the first quadrant is given by
$$\dfrac{1}{2} (b-a) \times 2a = a(b-a)$$
Here, $a = \dfrac{9}{4}$ and $b = 4$. Hence, the required area is $a(b-a) = \dfrac{9}{4} \times \dfrac{7}{4} = \dfrac{63}{16}$.
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Community Answer
The area bounded by the curve y2 = 9x and the lines x = 1, x = 4 and y...
Required area = 
= 2 (43/2 - 1) = 2 ( 8 - 1) = 14.
= 2 (43/2 - 1) = 2 ( 8 - 1) = 14.
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The area bounded by the curve y2 = 9x and the lines x = 1, x = 4 and y = 0 in the first quadrant isa)7b)14c)28d)14/3Correct answer is option 'B'. Can you explain this answer?
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