সৃজনশীলঃ ‘f(x)=x+1,g(x)=ex.…’ বরিশাল বোর্ড ২০২৫ – যোগজীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

দেওয়া আছে, $f\left(x\right)=x+1$ এবং $g\left(x\right)=e^x$

এখন, $\int \frac{f(x-1)g\left(x\right)}{\left\{f\right(x){\}}^2}dx$

$=\int \frac{x{e}^x}{(x+1{)}^2}dx$

$=\int e^x\left[\frac{x+1-1}{(x+1{)}^2}\right]dx$

$=\int e^x[\frac{x+1}{(x+1{)}^2}-\frac{1}{(x+1{)}^2}]dx$

$=\int e^x[\frac{1}{x+1}-\frac{1}{(x+1{)}^2}]dx$

$=\int e^x[\frac{1}{x+1}+\frac{d}{dx}\left(\frac{1}{x+1}\right)]dx$

$=\frac{e^x}{x+1}+c$

$[\because \int e^x\left(f\right(x)+f^{\prime }(x\left)\right)dx=e^{x}f(x)+c]$

$\therefore \int \frac{f(x-1)g\left(x\right)}{\left\{f\right(x){\}}^2}dx=\frac{e^x}{x+1}+c$

নির্ণেয় যোগজফল : $\frac{e^x}{x+1}+c$

ধরি, $x=4\sin \theta$

বা, $\mathrm{dx}=4\cos \theta d\theta$

দেওয়া আছে, $f\left(x\right)=x+1$

$\therefore {\int }_{-2}^2\left\{f\right(x-1){\}}^2\sqrt{16-\left\{f\right(x-1){\}}^2}dx$

$={\int }_{-2}^2{x}^2\sqrt{16-x^2}dx$

$={\int }_{-\frac{\pi }{6}}^{\frac{\pi }{6}}16{\sin }^2\theta \sqrt{16-16{\sin }^2\theta }\cdot 4\cos \theta d\theta$

$={\int }_{-\frac{\pi }{6}}^{\frac{\pi }{6}}16{\sin }^2\theta \times 4\sqrt{1-{\sin }^2\theta }\times 4\cos \theta d\theta$

$=64{\int }_{-\frac{\pi }{6}}^{\frac{\pi }{6}}(2\sin \theta \cos \theta {)}^{2}d\theta$$=32{\int }_{-\frac{\pi }{6}}^{\frac{\pi }{6}}2{\sin }^{2}2\theta d\theta$

$=32{\int }_{-\frac{\pi }{6}}^{\frac{\pi }{6}}(1-\cos 4\theta )d\theta$$[\because 2{\sin }^{2}A=1-\cos 2A]$

$=32{[\theta -\frac{\sin 4\theta }{4}]}_{-\frac{\pi }{6}}^{\frac{\pi }{6}}$

$=32[\{\frac{\pi }{6}-\frac{1}{4}\sin \left(\frac{4\pi }{6}\right)\}$$-\{-\frac{\pi }{6}-\frac{1}{4}\sin (-\frac{4\pi }{6})]$

$=32[\frac{\pi }{6}-\frac{1}{4}\sin \left(\frac{2\pi }{3}\right)$$+\frac{\pi }{6}-\frac{1}{4}\sin \left(\frac{2\pi }{3}\right)]$

$=32[\frac{\pi }{6}-\frac{1}{4}\cdot \frac{\sqrt{3}}{2}+\frac{\pi }{6}-\frac{1}{4}\frac{\sqrt{3}}{2}]$

$=32[\frac{2\pi }{6}-\frac{2\sqrt{3}}{8}]$

$=\frac{64\pi }{6}-\frac{64\sqrt{3}}{8}$$=\frac{32\pi }{3}-8\sqrt{3}$

নির্ণেয় মান : $\frac{32\pi }{3}-8\sqrt{3}$.