86886
The area of the region bounded by the curves \(f\) \((x)=\sin \pi x\) and \(X\)-axis is where \(x \in[-1,2]\)
1 \(8 \pi\)
2 \(\frac{8}{\pi}\)
3 \(\frac{6}{\pi}\)
4 \(6 \pi\)
Explanation:
(C) : since \(f(\mathrm{x})=\sin \pi \mathrm{x}\) is periodic function in nature. Hence, \(A=\int_{1}^{2} \sin \pi x d x\) put \(\pi \mathrm{x}=\mathrm{t} \Rightarrow \pi \mathrm{dx}=\mathrm{dt}\) \(\mathrm{A}=3 \int_{\pi}^{2 \pi} \frac{1}{\pi} \sin \mathrm{t} d \mathrm{t}\) \(=\frac{3}{\pi} \int_{\pi}^{0} \sin t d t=\frac{3}{\pi}[-\cos t]_{\pi}^{0}=\frac{3}{\pi}[-\cos 0+\cos (\pi)]\) \(A=\frac{6}{\pi}\) Since Area is not negative.
GUJCET-2017
Application of the Integrals
86789
The area of the region bounded by the line \(y=\) \(x-5\) and the \(x\)-axis between the ordinates \(x=3\) and \(x=7\)
86886
The area of the region bounded by the curves \(f\) \((x)=\sin \pi x\) and \(X\)-axis is where \(x \in[-1,2]\)
1 \(8 \pi\)
2 \(\frac{8}{\pi}\)
3 \(\frac{6}{\pi}\)
4 \(6 \pi\)
Explanation:
(C) : since \(f(\mathrm{x})=\sin \pi \mathrm{x}\) is periodic function in nature. Hence, \(A=\int_{1}^{2} \sin \pi x d x\) put \(\pi \mathrm{x}=\mathrm{t} \Rightarrow \pi \mathrm{dx}=\mathrm{dt}\) \(\mathrm{A}=3 \int_{\pi}^{2 \pi} \frac{1}{\pi} \sin \mathrm{t} d \mathrm{t}\) \(=\frac{3}{\pi} \int_{\pi}^{0} \sin t d t=\frac{3}{\pi}[-\cos t]_{\pi}^{0}=\frac{3}{\pi}[-\cos 0+\cos (\pi)]\) \(A=\frac{6}{\pi}\) Since Area is not negative.
GUJCET-2017
Application of the Integrals
86789
The area of the region bounded by the line \(y=\) \(x-5\) and the \(x\)-axis between the ordinates \(x=3\) and \(x=7\)
86886
The area of the region bounded by the curves \(f\) \((x)=\sin \pi x\) and \(X\)-axis is where \(x \in[-1,2]\)
1 \(8 \pi\)
2 \(\frac{8}{\pi}\)
3 \(\frac{6}{\pi}\)
4 \(6 \pi\)
Explanation:
(C) : since \(f(\mathrm{x})=\sin \pi \mathrm{x}\) is periodic function in nature. Hence, \(A=\int_{1}^{2} \sin \pi x d x\) put \(\pi \mathrm{x}=\mathrm{t} \Rightarrow \pi \mathrm{dx}=\mathrm{dt}\) \(\mathrm{A}=3 \int_{\pi}^{2 \pi} \frac{1}{\pi} \sin \mathrm{t} d \mathrm{t}\) \(=\frac{3}{\pi} \int_{\pi}^{0} \sin t d t=\frac{3}{\pi}[-\cos t]_{\pi}^{0}=\frac{3}{\pi}[-\cos 0+\cos (\pi)]\) \(A=\frac{6}{\pi}\) Since Area is not negative.
GUJCET-2017
Application of the Integrals
86789
The area of the region bounded by the line \(y=\) \(x-5\) and the \(x\)-axis between the ordinates \(x=3\) and \(x=7\)
86886
The area of the region bounded by the curves \(f\) \((x)=\sin \pi x\) and \(X\)-axis is where \(x \in[-1,2]\)
1 \(8 \pi\)
2 \(\frac{8}{\pi}\)
3 \(\frac{6}{\pi}\)
4 \(6 \pi\)
Explanation:
(C) : since \(f(\mathrm{x})=\sin \pi \mathrm{x}\) is periodic function in nature. Hence, \(A=\int_{1}^{2} \sin \pi x d x\) put \(\pi \mathrm{x}=\mathrm{t} \Rightarrow \pi \mathrm{dx}=\mathrm{dt}\) \(\mathrm{A}=3 \int_{\pi}^{2 \pi} \frac{1}{\pi} \sin \mathrm{t} d \mathrm{t}\) \(=\frac{3}{\pi} \int_{\pi}^{0} \sin t d t=\frac{3}{\pi}[-\cos t]_{\pi}^{0}=\frac{3}{\pi}[-\cos 0+\cos (\pi)]\) \(A=\frac{6}{\pi}\) Since Area is not negative.
GUJCET-2017
Application of the Integrals
86789
The area of the region bounded by the line \(y=\) \(x-5\) and the \(x\)-axis between the ordinates \(x=3\) and \(x=7\)
