Definite Integration MCQ Questions & Answers in Calculus | Maths

$$\int_0^{{t^2}} {xf\left( x \right)dx = \frac{2}{5}{t^5}\,\,\,\,\,\,\left( {{\text{Here, }}t > 0} \right)} $$
Differentiating both sides w.r.t. $$t$$ [Using Leibnitz theorem]
$$\eqalign{ & \Rightarrow {t^2}f\left( {{t^2}} \right) \times 2t - 0 = \frac{2}{5} \times 5{t^4} \cr & \Rightarrow f\left( {{t^2}} \right) = t \cr & {\text{Put }}t = \frac{2}{5}\,\,\,\,\,\,\,\,\,\, \Rightarrow f\left( {\frac{4}{{25}}} \right) = \frac{2}{5} \cr} $$

General term of the series $${\sum\limits_{n = 1}^{10} {\int\limits_{ - 2n - 1}^{ - 2n} {{{\sin }^{27}}x\,dx} } }$$       is
$$\eqalign{ & {I_1} = \int\limits_{ - 2n - 1}^{ - 2n} {{{\sin }^{27}}x\,dx} \cr & = \int\limits_{2n + 1}^{2n} {{{\sin }^{27}}\left( { - x} \right)\left( { - \,dx} \right)} \cr & = - \int\limits_{2n}^{2n + 1} {{{\sin }^{27}}x\,dx} \cr & = - {I_2} \cr} $$
Where $${I_2}$$ is general term of series
$$\eqalign{ & \sum\limits_{n = 1}^{10} {\int\limits_{2n}^{2n + 1} {{{\sin }^{27}}x\,dx} } \cr & {\text{So, }}{I_1} + {I_2} = 0{\text{ for all }}n \cr} $$

$$\int\limits_{ - 2}^3 {\left| {1 - {x^2}} \right|dx} = \int\limits_{ - 2}^3 {\left| {{x^2} - 1} \right|dx} $$
\[{\rm{Now }}\left| {{x^2} - 1} \right| = \left\{ \begin{array}{l} {x^2} - 1\,\,\,{\rm{if}}\,\,\,x \le - 1\\ 1 - {x^2}\,\,\,{\rm{if}}\,\,\, - 1 \le x \le 1\\ {x^2} - 1\,\,\,{\rm{if}}\,\,\,x \ge 1 \end{array} \right.\]
$$\eqalign{ & \therefore \,\,\,{\text{Integral is}} \cr & \int\limits_{ - 2}^{ - 1} {\left( {{x^2} - 1} \right)dx} + \int\limits_{ - 1}^1 {\left( {1 - {x^2}} \right)dx} + \int\limits_1^3 {\left( {{x^2} - 1} \right)dx} \cr & = \left[ {\frac{{{x^3}}}{3} - x} \right]_{ - 2}^{ - 1} + \left[ {x - \frac{{{x^3}}}{3}} \right]_{ - 1}^1 + \left[ {\frac{{{x^3}}}{3} - x} \right]_1^3 \cr & = \left( { - \frac{1}{3} + 1} \right) - \left( { - \frac{8}{3} + 2} \right) + \left( {2 - \frac{2}{3}} \right) + \left( {\frac{{27}}{3} - 3} \right) - \left( {\frac{1}{3} - 1} \right) \cr & = \frac{2}{3} + \frac{2}{3} + \frac{4}{3} + 6 + \frac{2}{3} \cr & = \frac{{28}}{3} \cr} $$

$$\eqalign{ & I = \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\frac{{\ln \left( {\cos \,x} \right)}}{{1 + {e^x}.{e^{\sin \,x}}}}} dx \cr & \Rightarrow I = \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\frac{{\ln \left( {\cos \,x} \right)}}{{1 + {e^{ - \left( {x + \sin \,x} \right)}}}}} dx \cr & \Rightarrow 2I = \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\frac{{\ln \left( {\cos \,x} \right)}}{{1 + {e^{x + \sin \,x}}}}\left( {1 + {e^{\left( {x + \sin \,x} \right)}}} \right)dx} \cr & \Rightarrow 2I = \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\ln \left( {\cos \,x} \right)dx} \cr & \Rightarrow 2I = 2\int\limits_0^{\frac{\pi }{2}} {\ln \left( {\cos \,x} \right)dx} \cr & \Rightarrow I = - \frac{\pi }{2}\ln \,2 \cr} $$

$$\eqalign{ & {I_n} = \int_0^{\frac{\pi }{2}} {{{\sin }^n}x\,dx} \cr & = \int_0^{\frac{\pi }{2}} {{{\sin }^{n - 1}}x.\sin \,x\,dx} \cr & = \left[ {{{\sin }^{n - 1}}x.\left( { - \cos \,x} \right)} \right]_0^{\frac{\pi }{2}} - \int_0^{\frac{\pi }{2}} { - \cos \,x.\left( {n - 1} \right){{\sin }^{n - 2}}x.\cos \,x\,dx} \cr & = \left( {n - 1} \right)\int_0^{\frac{\pi }{2}} {\left( {{{\sin }^{n - 2}}x - {{\sin }^n}x} \right)dx} \cr & \therefore \,\,\,n{I_n} = \left( {n - 1} \right){I_{n - 2}} \cr & \therefore \,\,\,{I_n} = \frac{{n - 1}}{n}{I_{n - 2}} \cr & \therefore \,\,\,{I_8} = \frac{7}{8}{I_6} = \frac{7}{8}.\frac{5}{6}.\frac{3}{4}.\frac{1}{2}{I_0} = \frac{{1.3.5.7}}{{1.2.3.4}}.\frac{1}{{{2^4}}}.\frac{\pi }{2} \cr} $$

$$\eqalign{ & {\text{Let }}{I_n} = \int\limits_0^\infty {{x^n}{e^{ - ax}}} \cr & = \left[ {{x^n}.\frac{{{e^{ - ax}}}}{{ - a}}} \right]_0^\infty - \int\limits_0^\infty {n{x^{n - 1}}.\frac{{{e^{ - ax}}}}{{ - a}}dx} \cr & = - \frac{1}{a}\mathop {\lim }\limits_{x \to \infty } \frac{{{x^n}}}{{{e^{ax}}}} + \frac{n}{a}{I_{n - 1}} \cr & \therefore \,{I_n} = \frac{n}{a}{I_{n - 1}} \cr & = \frac{n}{a}.\frac{{n - 1}}{a}{I_{n - 2}} \cr & = \frac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{{{a^3}}}{I_{n - 3}} \cr & = \frac{{n!}}{{{a^n}}}\int\limits_0^\infty {{e^{ - ax}}} dx \cr & = \frac{{n!}}{{{a^{n + 1}}}} \cr} $$

$$\eqalign{ & {I_1} = \int\limits_0^1 {{2^{{x^2}}}} dx,\,{I_2} = \int\limits_0^1 {{2^{{x^3}}}} dx \cr & = {I_3} = \int\limits_0^1 {{2^{{x^2}}}} dx,\,{I_4} = \int\limits_0^1 {{2^{{x^3}}}} dx\,\forall \,0 < x < 1,\,{x^2} > {x^3} \cr & \Rightarrow \int\limits_0^1 {{2^{{x^2}}}} dx > \int\limits_0^1 {{2^{{x^3}}}} dx \cr & \Rightarrow {I_1} > {I_2} \cr} $$

$$\eqalign{ & {\text{Put }}{x^2} = z \cr & {\text{Then }}\int_1^4 {\frac{{2.{e^{\sin \,{x^2}}}}}{x}dx} = \int_1^{16} {\frac{{{e^{\sin \,z}}}}{z}dz} \cr & = \int_1^{16} {\frac{d}{{dz}}\left\{ {F\left( z \right)} \right\}dz} \cr & = \left[ {F\left( z \right)} \right]_1^{16} \cr & = F\left( {16} \right) - F\left( 1 \right) \cr} $$
$$\therefore \,F\left( {16} \right) = F\left( k \right),$$     from the question.
Hence, one of the possible values of $$k=16.$$

$$\mathop {\lim }\limits_{x \to 0} \frac{1}{{{x^3}}}\int_0^x {\frac{{t\,ln\left( {1 + t} \right)}}{{{t^4} + 4}}dt\,\,\,\,\,\left[ {\frac{0}{0}{\text{form}}} \right]} $$
Applying L’ Hospital’s rule, we get
$$\eqalign{ & \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{xln\left( {1 + x} \right)}}{{{x^4} + 4}}}}{{3{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{ln\left( {1 + x} \right)}}{x}.\frac{1}{{3\left( {{x^4} + 4} \right)}} \cr & = 1.\frac{1}{{12}} \cr & = \frac{1}{{12}} \cr} $$

$$\int\limits_2^x {g\left( t \right)dt = } \frac{{{x^2}}}{2} + \int\limits_x^2 {{t^2}g\left( t \right)dt} $$
Differentiating w.r.t. $$x,$$ we get
$$\eqalign{ & g\left( x \right) = x + \left( { - {x^2}\left( {g\left( x \right)} \right)} \right) \cr & \Rightarrow g\left( x \right) = \frac{x}{{1 + {x^2}}} \cr} $$

Clearly from graph, $$\alpha \, \in \left( { - \frac{1}{2},\,\frac{1}{2}} \right) - \left\{ 0 \right\}$$