area bounded by the curves y e xy e x and the straight line

Area bounded by the curves $$y = {e^x},\,y = {e^{ - x}}$$ and the straight line $$x = 1$$ is (in sq. units)

A. $$e + \frac{1}{e}$$

B. $$e + \frac{1}{e} + 2$$

C. $$e + \frac{1}{e} - 2$$

D. $$e - \frac{1}{e} + 2$$

Answer : $$e + \frac{1}{e} - 2$$

Solution :
Given curves are $$y = {e^x}{\text{ and }}y = {e^{ - x}}$$
Now, $${e^x} = {e^{ - x}} \Rightarrow x = 0$$
$$\therefore $$ Area
$$\eqalign{ & = A = \int\limits_0^1 {\left( {{e^x} - {e^{ - x}}} \right)dx} \cr & = \left( {{e^x} + {e^{ - x}}} \right)_0^1 \cr & = \left[ {\left( {e + {e^{ - 1}}} \right) - \left( {{e^0} + {e^{ - 0}}} \right)} \right] \cr & = e + \frac{1}{e} - 2 \cr} $$

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

View Answer

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. $$4$$

View Answer

Releted Question 3

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

A. $$9$$

B. $$\frac{{27}}{4}$$

C. $$36$$

D. $$18$$

View Answer

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. $$\frac{1}{3}$$

View Answer

Area bounded by the curves $$y = {e^x},\,y = {e^{ - x}}$$ and the straight line $$x = 1$$ is (in sq. units)

Releted MCQ Question on
Calculus >> Application of Integration

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-

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

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

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-

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Area bounded by the curves $$y = {e^x},\,y = {e^{ - x}}$$ and the straight line $$x = 1$$ is (in sq. units)

Releted MCQ Question on Calculus >> Application of Integration

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-

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

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

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-

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Releted MCQ Question on
Calculus >> Application of Integration

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