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f(x) Transforms to [f(x)]

If f x be a f...

Question

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:

A

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B

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C

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D

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Solution

The correct option is A

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Similar questions

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

\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)

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 bounded by the curve

y = f (x)

, x-axis and the ordinates

x = 1

x = b

(b - 1) sin (3 b + 4)

f (x)

Q. The area bounded by the curve

y = f (x)

x -

axis and the ordinates

x = 1

x = b

(b - 1) cos (3 b + 4)

f (x)

Q. The area bounded by the curve

y = f (x)

x

x = a

x = b

f x

be a continuous function such that the area bounded by the curve

y = f (x)

x

-axis and the two ordinates

x = 0

x = a

\frac{a^{2}}{2} + \frac{a}{2} sin a + \frac{π}{2} cos a

f (\frac{π}{2})

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