86820
Area bounded by the curve \(y=x^{3}\), the \(x\)-axis and the ordinates at \(x=-2\) and \(x=1\), is
1 -9 sq. units
2 \(-\frac{15}{4}\) sq. units
3 \(\frac{15}{4}\) sq.units
4 \(\frac{17}{4}\) sq.units
Explanation:
(D) : Given curve \(y=x^{3}\) and ordinates, \(\mathrm{x}=-2, \mathrm{x}=1\) At \(x=-2\) then \(y=(-2)^{3}=-8\) and \(\mathrm{x}=1\) then \(\mathrm{y}=(1)^{3}=1\) Required area \(\mathrm{A}=\mathrm{A}_{1}+\mathrm{A}_{2}\) \(A=\left|\int_{-2}^{0} y \cdot d x\right|+\int_{0}^{1} y \cdot d x\) \(\Rightarrow \mathrm{A}=\int_{-2}^{0}-\mathrm{x}^{3} \cdot \mathrm{dx}+\int_{0}^{1} \mathrm{x}^{3} \cdot \mathrm{dx}\) \(\mathrm{A}=\left[-\frac{\mathrm{x}^{4}}{4}\right]_{-2}^{0}+\left[\frac{\mathrm{x}^{4}}{4}\right]_{0}^{1} \Rightarrow \mathrm{A}=\frac{16}{4}+\frac{4}{4}=\frac{16+1}{4}=\frac{17}{4}\) sq.units
COMEDK-2015
Application of the Integrals
86821
Find the area of the region bounded by the parabola \(y^{2}=4 a x\), its axis and two ordinates \(x=5\) and \(x=8\)
86820
Area bounded by the curve \(y=x^{3}\), the \(x\)-axis and the ordinates at \(x=-2\) and \(x=1\), is
1 -9 sq. units
2 \(-\frac{15}{4}\) sq. units
3 \(\frac{15}{4}\) sq.units
4 \(\frac{17}{4}\) sq.units
Explanation:
(D) : Given curve \(y=x^{3}\) and ordinates, \(\mathrm{x}=-2, \mathrm{x}=1\) At \(x=-2\) then \(y=(-2)^{3}=-8\) and \(\mathrm{x}=1\) then \(\mathrm{y}=(1)^{3}=1\) Required area \(\mathrm{A}=\mathrm{A}_{1}+\mathrm{A}_{2}\) \(A=\left|\int_{-2}^{0} y \cdot d x\right|+\int_{0}^{1} y \cdot d x\) \(\Rightarrow \mathrm{A}=\int_{-2}^{0}-\mathrm{x}^{3} \cdot \mathrm{dx}+\int_{0}^{1} \mathrm{x}^{3} \cdot \mathrm{dx}\) \(\mathrm{A}=\left[-\frac{\mathrm{x}^{4}}{4}\right]_{-2}^{0}+\left[\frac{\mathrm{x}^{4}}{4}\right]_{0}^{1} \Rightarrow \mathrm{A}=\frac{16}{4}+\frac{4}{4}=\frac{16+1}{4}=\frac{17}{4}\) sq.units
COMEDK-2015
Application of the Integrals
86821
Find the area of the region bounded by the parabola \(y^{2}=4 a x\), its axis and two ordinates \(x=5\) and \(x=8\)
86820
Area bounded by the curve \(y=x^{3}\), the \(x\)-axis and the ordinates at \(x=-2\) and \(x=1\), is
1 -9 sq. units
2 \(-\frac{15}{4}\) sq. units
3 \(\frac{15}{4}\) sq.units
4 \(\frac{17}{4}\) sq.units
Explanation:
(D) : Given curve \(y=x^{3}\) and ordinates, \(\mathrm{x}=-2, \mathrm{x}=1\) At \(x=-2\) then \(y=(-2)^{3}=-8\) and \(\mathrm{x}=1\) then \(\mathrm{y}=(1)^{3}=1\) Required area \(\mathrm{A}=\mathrm{A}_{1}+\mathrm{A}_{2}\) \(A=\left|\int_{-2}^{0} y \cdot d x\right|+\int_{0}^{1} y \cdot d x\) \(\Rightarrow \mathrm{A}=\int_{-2}^{0}-\mathrm{x}^{3} \cdot \mathrm{dx}+\int_{0}^{1} \mathrm{x}^{3} \cdot \mathrm{dx}\) \(\mathrm{A}=\left[-\frac{\mathrm{x}^{4}}{4}\right]_{-2}^{0}+\left[\frac{\mathrm{x}^{4}}{4}\right]_{0}^{1} \Rightarrow \mathrm{A}=\frac{16}{4}+\frac{4}{4}=\frac{16+1}{4}=\frac{17}{4}\) sq.units
COMEDK-2015
Application of the Integrals
86821
Find the area of the region bounded by the parabola \(y^{2}=4 a x\), its axis and two ordinates \(x=5\) and \(x=8\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Application of the Integrals
86820
Area bounded by the curve \(y=x^{3}\), the \(x\)-axis and the ordinates at \(x=-2\) and \(x=1\), is
1 -9 sq. units
2 \(-\frac{15}{4}\) sq. units
3 \(\frac{15}{4}\) sq.units
4 \(\frac{17}{4}\) sq.units
Explanation:
(D) : Given curve \(y=x^{3}\) and ordinates, \(\mathrm{x}=-2, \mathrm{x}=1\) At \(x=-2\) then \(y=(-2)^{3}=-8\) and \(\mathrm{x}=1\) then \(\mathrm{y}=(1)^{3}=1\) Required area \(\mathrm{A}=\mathrm{A}_{1}+\mathrm{A}_{2}\) \(A=\left|\int_{-2}^{0} y \cdot d x\right|+\int_{0}^{1} y \cdot d x\) \(\Rightarrow \mathrm{A}=\int_{-2}^{0}-\mathrm{x}^{3} \cdot \mathrm{dx}+\int_{0}^{1} \mathrm{x}^{3} \cdot \mathrm{dx}\) \(\mathrm{A}=\left[-\frac{\mathrm{x}^{4}}{4}\right]_{-2}^{0}+\left[\frac{\mathrm{x}^{4}}{4}\right]_{0}^{1} \Rightarrow \mathrm{A}=\frac{16}{4}+\frac{4}{4}=\frac{16+1}{4}=\frac{17}{4}\) sq.units
COMEDK-2015
Application of the Integrals
86821
Find the area of the region bounded by the parabola \(y^{2}=4 a x\), its axis and two ordinates \(x=5\) and \(x=8\)
