MCQ
$\int\limits_2^4 {(x(3 - x)(4 + x)(6 - x)(10 - x) + \sin x)} dx$ મેળવો.
- A$cos$ $2$ $+$ $cos$ $4$
- B$cos$ $2$ $-$ $cos$ $4$
- C$sin$ $2$ $+$ $sin$ $4$
- D$sin$ $2$ $-$ $sin$ $4$
$=\int_{2}^{4}((6-x)(3-(6-x))(4+(6-x))(6-(6-x))$
$\quad \ldots \ldots \ldots \int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x.$
$=\int_{2}^{4}((6-x)(x-3)(10-x) x(4+x)+\sin (6-x)) d x$
Adding equations (1) and (2), we get
$2 I =\int_{2}^{4}(\sin x+\sin (6-x)) d x$
$=(-\cos x+\cos (6-x))_{2}^{4}$
$=-\cos 4+\cos 2+\cos 2-\cos 4$
$=2(\cos 2-\cos 4)$
$\Rightarrow \quad I=\cos 2-\cos 4$
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વિધાન $2$:બિંદુ $A(1,0,7)$ અને $B(1,6,3)$ ને જોડતો રેખાખંડનેા રેખા $\frac{x}{1} = \frac{{y - 1}}{2} = \frac{{z - 2}}{3}$ દ્વિભાજક બને.