NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Application of the Integrals
86793
The area bounded by the line \(y=x, x\)-axis and coordinates \(x=-1\) and \(x=2\) is
1 \(3 / 2\) sq. units
2 \(5 / 2\) sq. units
3 2 sq. units
4 3 sq. units
Explanation:
(B) : Given line, \(\mathrm{y}=\mathrm{x}\) And coordinates, \(\mathrm{x}=-1, \mathrm{x}=2\) Total area, \(\quad \mathrm{A}=\mathrm{A}_{1}+\mathrm{A}_{2}\) \(\therefore \mathrm{A}=\int_{-1}^{0} \mathrm{x} . \mathrm{dx}+\int_{0}^{2} \mathrm{x} . \mathrm{dx}\) \(=\left[\frac{\mathrm{x}^{2}}{2}\right]_{-1}^{0}+\left[\frac{\mathrm{x}^{2}}{2}\right]_{0}^{2}=\left[\left.\frac{1}{2} \right\rvert\,\left(0^{2}-(-1)\right.\right.\) \(=\frac{1}{2}+\frac{1}{2} \times 4=\frac{1}{2}+\frac{4}{2}=\frac{5}{2}\) square units
Karnataka CET-2018
Application of the Integrals
86795
The area of the region bounded by the lines \(y=\) \(\mathrm{mx}, \mathrm{x}=1, \mathrm{x}=2\), and \(\mathrm{x}\) axis is \(6 \mathrm{sq}\). units, then ' \(\mathrm{m}\) ' is
86793
The area bounded by the line \(y=x, x\)-axis and coordinates \(x=-1\) and \(x=2\) is
1 \(3 / 2\) sq. units
2 \(5 / 2\) sq. units
3 2 sq. units
4 3 sq. units
Explanation:
(B) : Given line, \(\mathrm{y}=\mathrm{x}\) And coordinates, \(\mathrm{x}=-1, \mathrm{x}=2\) Total area, \(\quad \mathrm{A}=\mathrm{A}_{1}+\mathrm{A}_{2}\) \(\therefore \mathrm{A}=\int_{-1}^{0} \mathrm{x} . \mathrm{dx}+\int_{0}^{2} \mathrm{x} . \mathrm{dx}\) \(=\left[\frac{\mathrm{x}^{2}}{2}\right]_{-1}^{0}+\left[\frac{\mathrm{x}^{2}}{2}\right]_{0}^{2}=\left[\left.\frac{1}{2} \right\rvert\,\left(0^{2}-(-1)\right.\right.\) \(=\frac{1}{2}+\frac{1}{2} \times 4=\frac{1}{2}+\frac{4}{2}=\frac{5}{2}\) square units
Karnataka CET-2018
Application of the Integrals
86795
The area of the region bounded by the lines \(y=\) \(\mathrm{mx}, \mathrm{x}=1, \mathrm{x}=2\), and \(\mathrm{x}\) axis is \(6 \mathrm{sq}\). units, then ' \(\mathrm{m}\) ' is
86793
The area bounded by the line \(y=x, x\)-axis and coordinates \(x=-1\) and \(x=2\) is
1 \(3 / 2\) sq. units
2 \(5 / 2\) sq. units
3 2 sq. units
4 3 sq. units
Explanation:
(B) : Given line, \(\mathrm{y}=\mathrm{x}\) And coordinates, \(\mathrm{x}=-1, \mathrm{x}=2\) Total area, \(\quad \mathrm{A}=\mathrm{A}_{1}+\mathrm{A}_{2}\) \(\therefore \mathrm{A}=\int_{-1}^{0} \mathrm{x} . \mathrm{dx}+\int_{0}^{2} \mathrm{x} . \mathrm{dx}\) \(=\left[\frac{\mathrm{x}^{2}}{2}\right]_{-1}^{0}+\left[\frac{\mathrm{x}^{2}}{2}\right]_{0}^{2}=\left[\left.\frac{1}{2} \right\rvert\,\left(0^{2}-(-1)\right.\right.\) \(=\frac{1}{2}+\frac{1}{2} \times 4=\frac{1}{2}+\frac{4}{2}=\frac{5}{2}\) square units
Karnataka CET-2018
Application of the Integrals
86795
The area of the region bounded by the lines \(y=\) \(\mathrm{mx}, \mathrm{x}=1, \mathrm{x}=2\), and \(\mathrm{x}\) axis is \(6 \mathrm{sq}\). units, then ' \(\mathrm{m}\) ' is
86793
The area bounded by the line \(y=x, x\)-axis and coordinates \(x=-1\) and \(x=2\) is
1 \(3 / 2\) sq. units
2 \(5 / 2\) sq. units
3 2 sq. units
4 3 sq. units
Explanation:
(B) : Given line, \(\mathrm{y}=\mathrm{x}\) And coordinates, \(\mathrm{x}=-1, \mathrm{x}=2\) Total area, \(\quad \mathrm{A}=\mathrm{A}_{1}+\mathrm{A}_{2}\) \(\therefore \mathrm{A}=\int_{-1}^{0} \mathrm{x} . \mathrm{dx}+\int_{0}^{2} \mathrm{x} . \mathrm{dx}\) \(=\left[\frac{\mathrm{x}^{2}}{2}\right]_{-1}^{0}+\left[\frac{\mathrm{x}^{2}}{2}\right]_{0}^{2}=\left[\left.\frac{1}{2} \right\rvert\,\left(0^{2}-(-1)\right.\right.\) \(=\frac{1}{2}+\frac{1}{2} \times 4=\frac{1}{2}+\frac{4}{2}=\frac{5}{2}\) square units
Karnataka CET-2018
Application of the Integrals
86795
The area of the region bounded by the lines \(y=\) \(\mathrm{mx}, \mathrm{x}=1, \mathrm{x}=2\), and \(\mathrm{x}\) axis is \(6 \mathrm{sq}\). units, then ' \(\mathrm{m}\) ' is
