Definite Integration MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & {a_n} + {a_{n + 2}} = \int_0^{\frac{\pi }{4}} {\left( {{{\tan }^n}x + {{\tan }^{n + 2}}x} \right)dx,\,n \geqslant 2} \cr & = \int_0^{\frac{\pi }{4}} {{{\tan }^n}x} \,{\sec ^2}x\,dx \cr & = \left[ {\frac{{{{\tan }^{n + 1}}x}}{{n + 1}}} \right]_0^{\frac{\pi }{4}} \cr & = \frac{1}{{n + 1}} \cr} $$

$$\eqalign{ & {\text{Given,}} \cr & \int_0^1 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx} \cr & = \int_0^2 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx} \cr & = \int_0^1 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx} + \int_1^2 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx} \cr & \Rightarrow \int_1^2 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx} = 0 \cr} $$
Now we know that if $$\int_\alpha ^\beta {f\left( x \right)dx} = 0$$    then it means that $$f\left( x \right)$$  is $$+ve$$  on some part of $$\left( {\alpha ,\,\beta } \right)$$   and $$-ve$$  on other part of $$\left( {\alpha ,\,\beta } \right).$$
But here $${1 + {{\cos }^8}x}$$    is always $$+ve,$$
$$\therefore a{x^2} + bx + c$$    is $$+ve$$  on some part of $$\left[ {1,\,2} \right]$$  and $$-ve$$  on other Part $$\left[ {1,\,2} \right]$$
$$\therefore a{x^2} + bx + c = 0$$     has at least one root in $$\left( {1,\,2} \right)$$.
$$ \Rightarrow a{x^2} + bx + c = 0$$     has at least one root in $$\left( {0,\,2} \right)$$

$$\eqalign{ & {\text{Let }}I = \int\limits_{ - \frac{1}{2}}^{\frac{1}{2}} {\left( {\left[ x \right] + ln\left( {\frac{{1 + x}}{{1 - x}}} \right)} \right)} dx \cr & = \int\limits_{ - \frac{1}{2}}^{\frac{1}{2}} {\left[ x \right]} dx + \int\limits_{ - \frac{1}{2}}^{\frac{1}{2}} {ln\left( {\frac{{1 + x}}{{1 - x}}} \right)} dx \cr & = \int\limits_{ - \frac{1}{2}}^0 { - 1\,} dx + \int\limits_0^{\frac{1}{2}} {0\,} dx + 0 \cr & \left[ {\because \log \left( {\frac{{1 + x}}{{1 - x}}} \right){\text{ is an odd function}}} \right] \cr & = \left[ { - x} \right]_{ - \frac{1}{2}}^0 = 0 - \left( {\frac{1}{2}} \right) = - \frac{1}{2} \cr} $$