Application of Integration MCQ Questions & Answers in Calculus | Maths

Required area
$$\eqalign{ & = \int\limits_0^{2\pi } {\sin \,x\,dx} \cr & = \left. { - \cos \,x} \right|_0^{2\pi } \cr & = - \cos \,2\pi - \left( { - \cos \,0} \right) \cr & = - \cos \left( {\pi + \pi } \right) + 1 \cr & = - \left[ { - \cos \,\pi } \right] + 1 \cr & = + \cos \left( {\frac{\pi }{2} + \frac{\pi }{2}} \right) + 1 \cr & = \sin \frac{\pi }{2} + 1 \cr & = 1 + 1 \cr & = 2{\text{ sq}}{\text{. units}} \cr} $$

$$y = 1 + 4x - {x^2} = 5 - {\left( {x - 2} \right)^2}$$

We have $$\int\limits_0^{\frac{3}{2}} {\left( {1 + 4x - {x^2}} \right)} dx$$
$$\eqalign{ & = 2\int\limits_0^{\frac{3}{2}} {mx\,dx} \cr & = \frac{3}{2} + 2\left( {\frac{9}{4}} \right) - \frac{1}{3}\left( {\frac{{27}}{8}} \right) \cr & = m.\frac{9}{4} \cr} $$
On solving we get $$m = \frac{{13}}{6}$$

$$f\left( x \right) = {x^2} + bx - b\,;\,f'\left( x \right) = 2x + b\, \Rightarrow f'\left( 1 \right) = b + 2$$
Equation of tangent $$y - 1 = \left( {b + 2} \right)\left( {x - 1} \right)$$
Putting $$x = 0$$
$$ \Rightarrow y = 1 - b - 2 = - b - 1 > 0 \Rightarrow b < - 1$$
Putting $$y = 0$$
$$\eqalign{ & \Rightarrow x - 1 = - \frac{1}{{b + 2}} \cr & \Rightarrow x = \frac{{ - 1}}{{b + 2}} + 1 \cr & \Rightarrow x = \frac{{b + 1}}{{b + 2}} > 0 \cr & \Rightarrow b < - 2{\text{ or }}b > - 1 \cr} $$
Combining, the two conditions $$ = b < - 2$$
$$\eqalign{ & {\text{Now, }} \cr & \frac{1}{2}\left| { - b - 1} \right|\left| {\frac{{b + 1}}{{b + 2}}} \right| = 2 \cr & \Rightarrow {\left( {b + 1} \right)^2} = 4\left| {b + 2} \right| \cr & \Rightarrow {\left( {b + 1} \right)^2} = - 4b - 8 \cr & \Rightarrow {\left( {b + 3} \right)^2} = 0 \cr & \Rightarrow b = - 3\,\,{\text{follows the condition}}\,b < - 2 \cr} $$

$$\eqalign{ & I = \int_{\frac{\pi }{2}}^{\frac{{5\pi }}{6}} {\left[ {2\sin \,x} \right]dx} + \int_{\frac{{5\pi }}{6}}^\pi {\left[ {2\sin \,x} \right]dx} + \int_\pi ^{\frac{{7\pi }}{6}} {\left[ {2\sin \,x} \right]dx} + \int_{\frac{{7\pi }}{6}}^{\frac{{3\pi }}{2}} {\left[ {2\sin \,x} \right]dx} \cr & \,\,\,\,\, = \int_{\frac{\pi }{2}}^{\frac{{5\pi }}{6}} {1\,dx} + \int_{\frac{{5\pi }}{6}}^\pi {0\,dx} + \int_\pi ^{\frac{{7\pi }}{6}} { - 1\,dx} + \int_{\frac{{7\pi }}{6}}^{\frac{{3\pi }}{2}} { - 2\,dx} \cr & \,\,\,\,\, = \frac{{5\pi }}{6} - \frac{\pi }{2} - \left( {\frac{{7\pi }}{6} - \pi } \right) - 2\left( {\frac{{3\pi }}{2} - \frac{{7\pi }}{6}} \right) \cr & \,\,\,\,\, = - \frac{\pi }{2} \cr} $$

$$\eqalign{ & {\text{We have }}\int\limits_0^2 {f\left( x \right)dx} = \frac{3}{4}; \cr & {\text{Now, }} \cr & \int\limits_0^2 {x\,f'\left( x \right)dx} = x\int\limits_0^2 {f'\left( x \right)dx} - \int\limits_0^2 {f\left( x \right)dx} \cr & = \left[ {x\,f\left( x \right)} \right]_0^2 - \frac{3}{4} = 2f\left( 2 \right) - \frac{3}{4} \cr & = 0 - \frac{3}{4}\,\,\,\left[ {\because f\left( 2 \right) = 0} \right] \cr & = - \frac{3}{4} \cr} $$

Clearly, $$\int_1^2 {\left( {1 + {{\sin }^4}x} \right)} \left( {a{x^2} + bx + c} \right)dx = 0$$
As $$1 + {\sin ^4}x > 0$$    for all $$x\, \in \left[ {1,\,2} \right],$$   the above can be true only if $${a{x^2} + bx + c}$$     changes sign in [1, 2]. So, $$a{x^2} + bx + c = 0$$     has at least one real root between 1 and 2.

$${\text{Area}} = \frac{1}{2}.1.1 + \frac{1}{2}.5.5.\frac{{26}}{2} = 13$$

$$\eqalign{ & \phi \left( x \right) = \int_0^{{x^2}} {\left( {t - 1} \right)dt} - \int_0^x {\left( {t - 1} \right)dt} \cr & \therefore \phi '\left( x \right) = \left( {{x^2} - 1} \right).2x - \left( {x - 1} \right) = 2{x^3} - 3x + 1 \cr & \therefore \phi '\left( x \right) = \left( {x - 1} \right)\left( {2{x^2} + 2x - 1} \right) \cr & = 2\left( {x - 1} \right)\left( {{x^2} + x - \frac{1}{2}} \right) \cr & = 2\left( {x - 1} \right)\left\{ {{{\left( {x + \frac{1}{2}} \right)}^2} - \frac{3}{4}} \right\} \cr} $$
In $$1 \leqslant x \leqslant 2,\,\phi '\left( x \right) \geqslant 0.$$      So, $$\phi \left( x \right)$$  is m.i. in [1, 2]. Hence, $$\phi \left( 2 \right)$$  is the greatest value. Now, $$\phi \left( 2 \right) = \int_2^4 {\left( {t - 1} \right)dt} = \left[ {\frac{{{t^2}}}{2} - t} \right]_2^4 = \left( {8 - 4} \right) - \left( {2 - 2} \right) = 4\,$$

Required area $$\left( {OAB} \right)$$
$$\eqalign{ & = \int\limits_{1 - e}^0 {\ln \left( {x + e} \right)dx} \cr & = \left[ {x\,\ln \left( {x + e} \right) - \int {\frac{1}{{x + e}}x\,dx} } \right]_0^1 \cr & = 1 \cr} $$

$$\eqalign{ & {\text{Let }}x - \left[ x \right] = z.{\text{ In }}0 \leqslant x \leqslant \frac{\pi }{4},\,\left[ x \right] = 0.\,\,{\text{So }}x = z. \cr & \therefore I = \int_0^{\frac{\pi }{4}} {\sin \,z\,dz} \,\, = \left[ { - \cos \,z} \right]_0^{\frac{\pi }{4}}\,\, = 1 - \frac{1}{{\sqrt 2 }} \cr} $$