Application of Integration MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & I = \int_0^a {\frac{{dx}}{{1 + {e^{f\left( x \right)}}}}} = \int_0^a {\frac{{dx}}{{1 + {e^{f\left( {a - x} \right)}}}}} = \int_0^a {\frac{{dx}}{{1 + {e^{ - f\left( x \right)}}}}} \left( {{\text{from the question}}} \right) \cr & \,\,\,\,\, = \int_0^a {\frac{{{e^{f\left( x \right)}}dx}}{{{e^{f\left( x \right)}} + 1}}} \cr & \therefore I + I = \int_0^a {\frac{{dx}}{{1 + {e^{f\left( x \right)}}}}} + \int_0^a {\frac{{{e^{f\left( x \right)}}}}{{1 + {e^{f\left( x \right)}}}}} dx \cr & \therefore 2I = \int_0^a {dx} = a \cr} $$

$$\eqalign{ & f'\left( x \right) = x\left( {{e^x} - 1} \right){\left( {x - 2} \right)^3}.{\left( {x - 3} \right)^5} \cr & f'\left( {0 - \in } \right) = \left( {0 - \in } \right)\left( {{e^{0 - \in }} - 1} \right){\left( {0 - \in - 2} \right)^3}{\left( {0 - \in - 3} \right)^5} = \left( { + {\text{ve}}} \right) \cr & f'\left( {0 + \in } \right) = \left( {0 + \in } \right)\left( {{e^{0 + \in }} - 1} \right){\left( {0 + \in - 2} \right)^3}{\left( {0 + \in - 3} \right)^5} = \left( { + {\text{ve}}} \right) \cr & f'\left( {1 - \in } \right) = \left( {1 - \in } \right)\left( {{e^{1 - \in }} - 1} \right){\left( {1 - \in - 2} \right)^3}{\left( {1 - \in - 3} \right)^5} = \left( { + {\text{ve}}} \right) \cr & f'\left( {1 + \in } \right) = \left( {1 + \in } \right)\left( {{e^{1 + \in }} - 1} \right){\left( {1 + \in - 2} \right)^3}{\left( {1 + \in - 3} \right)^5} = \left( { + {\text{ve}}} \right) \cr & f'\left( {2 - \in } \right) = \left( {2 - \in } \right)\left( {{e^{2 - \in }} - 1} \right){\left( {2 - \in - 2} \right)^3}{\left( {2 - \in - 3} \right)^5} = \left( { + {\text{ve}}} \right) \cr & f'\left( {2 + \in } \right) = \left( {2 + \in } \right)\left( {{e^{2 + \in }} - 1} \right){\left( {2 + \in - 2} \right)^3}{\left( {2 + \in - 3} \right)^5} = \left( { - {\text{ve}}} \right) \cr & {\text{so, a local maximum at }}x = 2 \cr & f'\left( {3 - \in } \right) = \left( {3 - \in } \right)\left( {{e^{3 - \in }} - 1} \right){\left( {3 - \in - 2} \right)^3}{\left( {3 - \in - 3} \right)^5} = \left( { - {\text{ve}}} \right) \cr & f'\left( {3 + \in } \right) = \left( {3 + \in } \right)\left( {{e^{3 + \in }} - 1} \right){\left( {3 + \in - 2} \right)^3}{\left( {3 + \in - 3} \right)^5} = \left( { + {\text{ve}}} \right); \cr & {\text{so, a local minimum at }}x = 3 \cr} $$

$$\eqalign{ & I = \int_{ - 2}^2 {x\,dx} - \int_{ - 2}^2 {\left[ x \right]dx} \cr & = 0 - \left\{ {\int_{ - 2}^{ - 1} {\left[ x \right]dx + \int_{ - 1}^0 {\left[ x \right]dx} + \int_0^1 {\left[ x \right]dx} + \int_1^2 {\left[ x \right]dx} } } \right\} \cr & = - \int_{ - 2}^{ - 1} { - 2\,dx} - \int_{ - 1}^0 { - 1\,dx} - \int_0^1 {0\,dx} - \int_1^2 {1\,dx} \cr & = 2\left| x \right|_{ - 2}^{ - 1} + \left| x \right|_{ - 1}^0 - 0 - \left| x \right|_1^2 \cr & = 2\left( { - 1 + 2} \right) + \left( {0 + 1} \right) - \left( {2 - 1} \right) \cr & = 2 \cr} $$

Clearly, the integrand is an odd function. So, $$I=0.$$

$$\eqalign{ & I = \int_0^\pi {\frac{{\sin \,n\left( {\pi - x} \right)}}{{\sin \left( {\pi - x} \right)}}dx = \int_0^\pi {\frac{{{{\left( { - 1} \right)}^{n - 1}}\sin \,nx}}{{\sin \,x}}dx} = {{\left( { - 1} \right)}^{n - 1}}I} \cr & \therefore {\text{ if }}n\,{\text{is even, }}I = - I\,\, \Rightarrow I = 0 \cr} $$

Area of the region is given by
$$A = \int\limits_{ - 1}^3 {\left[ {\left( {y + 1} \right) - \left( {\frac{{{y^2} - 1}}{2}} \right)} \right]dy} = \frac{{16}}{3}$$

$$\eqalign{ & \because {\text{Curve is given as}}: \cr & y = {x^2} - 1 \cr & \Rightarrow \frac{{dy}}{{dx}} = 2x \cr & \Rightarrow {\left( {\frac{{dy}}{{dx}}} \right)_{\left( {2,\,3} \right)}} = 4 \cr & \therefore {\text{Equation of tangent at }}\left( {2,{\text{ }}3} \right) \cr & \left( {y - 3} \right) = {\text{4}}\left( {x - 2} \right) \cr & \Rightarrow y = 4x - 5 \cr & {\text{but }}x = 0 \cr & \Rightarrow y = - 5 \cr} $$
Here the curve cuts $$y$$-axis
$$\eqalign{ & \therefore {\text{Required area:}} \cr & = \frac{1}{4}\int\limits_{ - 5}^3 {\left( {y + 5} \right)dy} - \int\limits_{ - 1}^3 {\sqrt {y + 1} \,dy} \cr & = \frac{1}{4}\left[ {\frac{{{y^2}}}{2} + 5y} \right]_{ - 5}^3\frac{{ - 2}}{3}\left[ {{{\left( {y + 1} \right)}^{\frac{3}{2}}}} \right]_{ - 1}^3 \cr & = \frac{1}{4}\left[ {\frac{9}{2} + 15 - \frac{{25}}{2} + 25} \right] - \frac{2}{3}\left[ {{4^{\frac{3}{2}}} - 0} \right] \cr & = \frac{{32}}{4} - \frac{{16}}{3} \cr & = \frac{8}{3}{\text{ sq}}{\text{. units}} \cr} $$

First we draw each curve as separate graph

NOTE : Graph of $$y = \left| {f\left( x \right)} \right|$$   can be obtained from the graph of the curve $$y = f\left( x \right)$$   by drawing the mirror image of the portion of the graph below $$x$$-axis, with respect to $$x$$-axis.
Clearly the bounded area is as shown in the following figure.

$$\eqalign{ & {\text{Required area}}\, = 4\int\limits_0^1 {\left( { - \ell nx} \right)dx} \cr & = - 4\left[ {x\,\ell n\,x - x} \right]_0^1 \cr & = 4\,{\text{sq}}{\text{. units}} \cr} $$

$$ATQ\,\int_1^b {f\left( x \right)dx} = \left( {b - 1} \right)\sin \left( {3b + 4} \right)$$
Differentiating both sides w.r.t $$b,$$  we get
$$\eqalign{ & f\left( b \right) = 3\left( {b - 1} \right)\cos \left( {3b + 4} \right) + \sin \left( {3b + 4} \right) \cr & \Rightarrow f\left( x \right) = 3\left( {x - 1} \right)\cos \left( {3x + 4} \right) + \sin \left( {3x + 4} \right) \cr} $$

$$\eqalign{ & I = \int_{ - 1}^0 {f\left( x \right)dx} + \int_0^1 {f\left( x \right)dx} \cr & \,\,\,\,\, = \int_{ - 1}^0 {\left( {\left| x \right| + 1} \right)dx} + \int_0^1 {\left( {1 + {{\left| x \right|}^2}} \right)dx} \cr & \,\,\,\,\, = \int_{ - 1}^0 {\left( { - x + 1} \right)dx} + \int_0^1 {\left( {1 + {x^2}} \right)dx} \cr & \,\,\,\,\, = \left[ { - \frac{{{x^2}}}{2} + x} \right]_{ - 1}^0 + \left[ {x + \frac{{{x^3}}}{3}} \right]_0^1 \cr & \,\,\,\,\, = - \left( { - \frac{1}{2} - 1} \right) + \left( {1 + \frac{1}{3}} \right) \cr & \,\,\,\,\, = \frac{3}{2} + \frac{4}{3} \cr & \,\,\,\,\, = \frac{{17}}{6} \cr} $$