Question

The value of $$\int_{ - 1}^1 {\max \left\{ {2 - x,\,2,\,1 + x} \right\}} $$      is :

A. 4
B. $$\frac{9}{2}$$  
C. 2
D. none of these
Answer :   $$\frac{9}{2}$$
Solution :
$$\eqalign{ & I = \int_{ - 1}^0 {\max \left\{ {2 - x,\,2,\,1 + x} \right\}} dx + \int_0^1 {\max \left\{ {2 - x,\,2,\,1 + x} \right\}} dx \cr & \,\,\,\,\, = \int_{ - 1}^0 {\left( {2 - x} \right)dx} + \int_0^1 {2\,dx} \cr & \,\,\,\,\, = \left[ {2x - \frac{{{x^2}}}{2}} \right]_{ - 1}^0 + 2\left[ x \right]_0^1 \cr & \,\,\,\,\, = - \left( { - 2 - \frac{1}{2}} \right) + 2 \cr & \,\,\,\,\, = \frac{9}{2} \cr} $$

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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