Question

If $$f\left( { - x} \right) + f\left( x \right) = 0$$     then $$\int_a^x {f\left( t \right)dt} $$   is :

A. an odd function
B. an even function  
C. a periodic function
D. none of these
Answer :   an even function
Solution :
$$\eqalign{ & {\text{Let }}\phi \left( x \right) = \int_a^x {f\left( t \right)dt.\,{\text{Then }}\phi } \left( { - x} \right) = \int_a^{ - x} {f\left( t \right)dt} \cr & \therefore \phi \left( { - x} \right) = \int_a^x {f\left( t \right)dt} + \int_x^{ - x} {f\left( t \right)dt} \cr & = \phi \left( x \right) + \int_x^0 {f\left( t \right)dt} + \int_0^{ - x} {f\left( t \right)dt} \cr & = \phi \left( x \right) - \int_0^x {f\left( t \right)dt} + \int_0^x { - f\left( { - z} \right)dz,{\text{ using }}t} = - z \cr & = \phi \left( x \right) - \int_0^x {\left\{ {f\left( t \right) + f\left( { - t} \right)} \right\}dt} \cr & = \phi \left( x \right)\,\,\,\,\,\,\,\,\,\,\,\,\left( {\because f\left( t \right) + f\left( { - t} \right) = 0,{\text{ from the question}}} \right) \cr & \therefore \,\,\,\phi \left( x \right){\text{ is even}}{\text{.}} \cr} $$

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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Application of Integration


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