(i) [গণিত] এর মান নির্ণয় কর। (ii) [গণিত] এর মাन নির্ণয় কর।
উত্তর: [গণিত] (ii) [গণিত] ধরি, [গণিত] [চিত্র] [গণিত]
HSC Higher Math 1st Paper Chapter 10: যোগজীকরণ CQ — Prosthuti প্রশ্নব্যাংক
HSC · Higher Math 1st Paper · Chapter 10: যোগজীকরণ · সৃজনশীল
(i) এর মান নির্ণয় কর। (ii) এর মাन নির্ণয় কর।
উত্তর
\begin{aligned} & \text{ (i) } \int_0^2 f(x) \tan^{-1}(x-2) d x=\int_0^2(x-2) \tan^{-1}(x-2) d x \\ & \quad \therefore \int(x-2) \tan^{-1}(x-2) d x \\ & =\tan^{-1}(x-2) \int(x-2) d x-\left[\frac{d}{d x} \tan^{-1}(x-2) \mathrm{o}(x-2) d x\right] d x \\ & =\left(\frac{x^2}{2}-2 x\right) \tan^{-1}(x-2)-\int \frac{1}{1+(x-2}^2} \cdot 1 \cdot\left(\frac{x^2}{2}-2 x\right) d x \\ & =\frac{1}{2}\left(x^2-4 x\right) \tan^{-1}(x-2)-\frac{1}{2} \int \frac{x^2-4 x}{x^2-4 x+5} d x \\ & =\frac{1}{2}\left(x^2-4 x\right) \tan^{-1}(x-2)-\frac{1}{2} \int \frac{x^2-4 x+5-5}{x^2-4 x+5} d x \\ & =\frac{1}{2}\left(x^2-4 x\right) \tan^{-1}(x-2)-\frac{1}{2} \int\left(1-\frac{5}{(x-2}^2+1^2}\right) d x \\ & =\frac{1}{2}\left(x^2-4 x\right) \tan^{-1}(x-2)-\frac{1}{2} x+\frac{1}{2} \cdot 5 \tan^{-1}(x-2)+c \\ & =\frac{1}{2}\left\{\left(x^2-4 x+5\right) \tan^{-1}(x-2)-x\right\}+c \\ & \therefore \int_0^2 f(x) \tan^{-1}(x-2) d x \\ & =\frac{1}{2}\left[\left(x^2-4 x+5\right) \tan^{-1}(x-2)-x\right]_0^2 \\ & =\frac{1}{2}\left[\left\{\left(2^2-4 \times 2+5\right) \tan^{-1}(2-2)-2\right\}\right. \\ & \left.-\left\{\left(0^2-4.0+5\right) \tan^{-1}(0-2)-0\right\}\right] \\ & =\frac{1}{2}\left\{0-2-5 \tan^{-1}(-2)\right\}\end{aligned}
(ii)
ধরি,
