Question

$$\int_0^{\frac{\pi }{4}} {\sin \,x\,d\left( {x - \left[ x \right]} \right)} $$     is equal to :

A. $$\frac{1}{2}$$
B. $$1 - \frac{1}{{\sqrt 2 }}$$  
C. 1
D. none of these
Answer :   $$1 - \frac{1}{{\sqrt 2 }}$$
Solution :
$$\eqalign{ & {\text{Let }}x - \left[ x \right] = z.{\text{ In }}0 \leqslant x \leqslant \frac{\pi }{4},\,\left[ x \right] = 0.\,\,{\text{So }}x = z. \cr & \therefore I = \int_0^{\frac{\pi }{4}} {\sin \,z\,dz} \,\, = \left[ { - \cos \,z} \right]_0^{\frac{\pi }{4}}\,\, = 1 - \frac{1}{{\sqrt 2 }} \cr} $$

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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