Question

Let $$f\left( x \right)$$  be a given integrable function such that $$f\left( {x + k} \right) = f\left( x \right)$$    for all $$x\, \in \,R.$$   Then $$\int_a^{a + k} {f\left( x \right)dx} $$    depends for its value on :

A. $$a$$ only
B. $$k$$ only  
C. both $$a$$ and $$k$$
D. neither $$a$$ nor $$k$$
Answer :   $$k$$ only
Solution :
$$\eqalign{ & \int_a^{a + k} {f\left( x \right)} dx = \int_0^{a + k} {f\left( x \right)} dx - \int_0^a {f\left( x \right)} dx \cr & = \int_0^k {f\left( x \right)} dx + \int_k^{a + k} {f\left( x \right)} dx - \int_0^a {f\left( x \right)} dx \cr & {\text{For the second integral, put }}x = z + k \cr & {\text{Then }}\int_k^{a + k} {f\left( x \right)} dx = \int_0^a {f\left( {z + k} \right)} dz = \int_0^a {f\left( z \right)} dz \cr & \therefore \int_a^{a + k} {f\left( x \right)} dx = \int_0^k {f\left( x \right)} dx + \int_0^a {f\left( x \right)} dx - \int_0^a {f\left( x \right)} dx \cr & = \int_0^k {f\left( x \right)} dx \cr & {\text{which depends on }}k{\text{ but not on }}a. \cr} $$

Releted MCQ Question on
Calculus >> Application of Integration

Releted Question 1

The area bounded by the curves $$y = f\left( x \right),$$   the $$x$$-axis and the ordinates $$x = 1$$  and $$x = b$$  is $$\left( {b - 1} \right)\sin \left( {3b + 4} \right).$$     Then $$f\left( x \right)$$  is-

A. $$\left( {x - 1} \right)\cos \left( {3x + 4} \right)$$
B. $$\sin \,\left( {3x + 4} \right)$$
C. $$\sin \,\left( {3x + 4} \right) + 3\left( {x - 1} \right)\cos \left( {3x + 4} \right)$$
D. none of these
Releted Question 2

The area bounded by the curves $$y = \left| x \right| - 1$$   and $$y = - \left| x \right| + 1$$   is-

A. $$1$$
B. $$2$$
C. $$2\sqrt 2 $$
D. $$4$$
Releted Question 3

The area bounded by the curves $$y = \sqrt x ,\,2y + 3 = x$$    and $$x$$-axis in the 1st quadrant is-

A. $$9$$
B. $$\frac{{27}}{4}$$
C. $$36$$
D. $$18$$
Releted Question 4

The area enclosed between the curves $$y = a{x^2}$$   and $$x = a{y^2}\left( {a > 0} \right)$$    is 1 sq. unit, then the value of $$a$$ is-

A. $$\frac{1}{{\sqrt 3 }}$$
B. $$\frac{1}{2}$$
C. $$1$$
D. $$\frac{1}{3}$$

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