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Byju's Answer

Standard XII

Mathematics

Area between x=g(y) and y Axis

The area boun...

Question

The area bounded by the curve $y = f (x)$ , above the $x$ -axis, between $x = a$ and $x = b$ is:

\int_{f (a)}^{b} y d y

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\int_{a}^{f (b)} x d x

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\int_{a}^{b} x d y

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\int_{a}^{b} y d x

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Solution

The correct option is D $\int_{a}^{b} y d x$
We need to calculate the area bounded by $y = f (x)$ between $x = a$ and $x = b$
So, we need to find the area of the region inside the curve $y = f (x)$ having limit points $a$ to $b$ and it is given by
$\int_{a}^{b} f (x) d x$
$= \int_{a}^{b} y d x$
Hence, option D is correct.

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Similar questions

If f(x) be a function defined on the closed interval [a, b] and graph of the function f(x) is a curve above X-axis, the area bounded by the curve f(x) and the ordinates x = a, x = b and X-axis is:

Let y=f(x) be the given curve and

x = a

x = b

be two ordinates then area bounded by the curve

y = f (x)

, the axis of x between the ordinates

x = a

x = b

, is given by definite integral

\int_{a}^{b} y d x

\int_{a}^{b} f (x) d x

and the area bounded by the curve

x = f (y)

, the axis of y & two abscissae

y = c

y = d

is given by

\int_{c}^{d} x d y

\int_{c}^{d} f (x) d y

. Again if we consider two curves

y = f (x)

y = g (x)

where

f (x) \geq g (x)

in the interval [a, b] where

x = a

x = b

are the points of intersection of these two curves Shown by the graph given
Then area bounded by these two curves is given by

\int_{a}^{b} [f (x) - g (x)] d x

On the basis of above information answer the following questions.

The area bounded by the curves

y = \sqrt{x}

x = 2 y + 3

and x-axis (in the first quadrant) is equal to (in square units)

Q. The area above x-axis, bounded by the line

x = 4

and the curve

y = f (x)

, where

f (x) = x^{2}, 0 \leq x \leq 1

and

f (x) = \sqrt{x}, x \geq 1

, is

Q. I: The area bounded by the curve

y = {log}_{e} x, X

-axis and the line

x = e

is 1 sq. units.
II: The area bounded by

X

-axis, the curve

y = f (x)

and the lines

x = 1, x = b

\sqrt{b^{2} + 1} - \sqrt{2}

for all

b > 1

then

f (x) = \frac{x}{\sqrt{x^{2} + 1}}

Which of the above statement is correct?

Q. Solve:

x d x + y d y + \frac{x d y - y d x}{x^{2} + y^{2}} = 0

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The area bounded by the curve y=f(x), above the x-axis, between x=a and x=b is:

The correct option is D ∫baydxWe need to calculate the area bounded by y=f(x) between x=a and x=bSo, we need to find the area of the region inside the curve y=f(x) having limit points a to b and it is given by∫baf(x)dx=∫baydxHence, option D is correct.

The area bounded by the curve $y = f (x)$ , above the $x$ -axis, between $x = a$ and $x = b$ is: