Match the following :

Question

Professor PC Thomas Classes · Answer

(A) $I=\int_{-2}^2\left(\alpha x^3+\beta x+\gammaight) d x$ 
$\alpha x^3+\beta x$ is an odd function 
$ I=0+2 \int_0^2 \gamma d x=2.2 \gamma=4 \gamma $
 (B) $I=\frac{1}{2} \int_0^1 2 \sin \alpha x \sin \beta x d x=\frac{1}{2} \int_0^1(\cos (\alpha-\beta) x-\cos (\alpha+\beta) x) d x$ 
$ =\frac{1}{2}\left[\frac{\sin (\alpha-\beta) x}{\alpha-\beta}-\frac{\sin (\alpha+\beta) x}{\alpha+\beta}ight]_0^1=\frac{1}{2}\left[\frac{\sin (\alpha-\beta)}{\alpha-\beta}-\frac{\sin (\alpha+\beta)}{\alpha+\beta}ight] $
 Also, $2 \alpha=	an \alpha$ and $2 \beta=	an \beta \Rightarrow 2(\alpha-\beta)=	an \alpha-	an \beta$ and $2(\alpha+\beta)=	an \alpha+	an \beta$ 
$ 2(\alpha-\beta)=\frac{\sin (\alpha-\beta)}{\cos \alpha \cos \beta} 	ext { and } 2(\alpha+\beta)=\frac{\sin (\alpha+\beta)}{\cos \alpha \cos \beta} $
 Substituting these values, we get : $I=(\cos \alpha \cos \beta)-(\cos \alpha \cos \beta)=0$. \ (C) $f(x+\alpha)+f(x)=0 \Rightarrow f(x+2 \alpha)+f(x+\alpha)=0 \Rightarrow f(x+2 \alpha)=f(x)$ 
$\Rightarrow f(x)$ is periodic with period $2 \alpha \Rightarrow \int_\beta^{\beta+2 \gamma \alpha} f(x) d x=\gamma \int_0^{2 \alpha} f(x) d x$.