86856
The area of the curve bounded by \(x^{2}+y^{2}=4\) and \(x+y=4\) is
1 \(\pi\)
2 \(\pi-8\)
3 \(\pi+8\)
4 \(8-\pi\)
Explanation:
(D) : Area of curve bounded by \(x^{2}+y^{2}=4\) and \(x+y=4\) Now, area of \(\triangle \mathrm{AOB}\), \(\frac{1}{2} \times 4 \times 4=8\) Area of quarter circle \(\left(\frac{1}{4} \pi(2)^{2}\right)=\pi\) Area of shaded region \(=\) Area of triangle - Area of quarter circle \(=8-\pi\)
AMU-2004
Application of the Integrals
86864
Area bounded by the curve \(x=0\) and \(x+2|y|=\) 1 is :
86857
The area bounded by the curve \(y=2 x^{4}-x^{2}\), the \(x\)-axis and the two ordinates corresponding to minimal of the function is
1 \(\frac{1}{40}\) square unit
2 \(\frac{7}{120}\) square unit
3 \(\frac{1}{24}\) square unit
4 None of the above
Explanation:
(B): Given the curve, \(y=2 x^{4}-x^{2}\) \(\frac{d y}{d x}=8 x^{3}-2 x=0=4 x^{3}-x=0=x\left(4 x^{2}-1\right)=0\) \(\therefore \mathrm{x}=0, \quad 4 \mathrm{x}^{2}=1 \Rightarrow \mathrm{x}^{2}=\frac{1}{4}\) \(\mathrm{x}= \pm \frac{1}{2}\) Now, area of curves, \(A=\int_{-1 / 2}^{1 / 2}\left(2 x^{4}-x^{2}\right) d x\) \(A=2\left[\int_{0}^{1 / 2}\left(2 x^{4}-x^{2}\right) d x\right]\) (By using even property) \(=2\left[\frac{2 \mathrm{x}^{5}}{5}-\frac{\mathrm{x}^{3}}{3}\right]_{0}^{1 / 2}=2\left|2 \times \frac{1}{32 \times 5}-\frac{1}{8 \times 3}\right|\) \(=2 \times\left|\frac{1}{16 \times 5}-\frac{1}{8 \times 3}\right|\) \(=2\left(\frac{24-80}{16 \times 5 \times 8 \times 3}\right)=\frac{56 \times 2}{16 \times 5 \times 8 \times 3}=\frac{7}{120}\) square unit
SCRA-2014
Application of the Integrals
86858
The line \(x=\frac{\pi}{4}\) divides the area of the region bounded by \(y=\sin x, y=\cos x\) and \(x\)-axis \(\left(0 \leq x \leq \frac{\pi}{2}\right)\) into two regions of areas \(A_{1}\) and \(\mathbf{A}_{2}\). Then \(\mathbf{A}_{1}, \mathbf{A}_{2}\) equals
86859
The area bounded by \(y=x^{2}+2, x\)-axis, \(x=1\) and \(x=2\) is
1 \(\frac{16}{3}\) sq unit
2 \(\frac{17}{3}\) sq unit
3 \(\frac{13}{3}\) sq unit
4 \(\frac{20}{3}\) sq unit
Explanation:
(C) : Given curve, \(\mathrm{y}=\mathrm{x}^{2}+2\) \(\therefore\) The required area is \(A=\int_{1}^{2} y \cdot d x=\int_{1}^{2}\left(x^{2}+2\right) d x=\int_{1}^{2} x^{2} d x+\int_{1}^{2} 2 \cdot d x\) \(=\left[\frac{\mathrm{x}^{3}}{3}\right]_{1}^{2}+2 \cdot[\mathrm{x}]_{1}^{2}=\frac{1}{3}(8-1)+2(2-1)\) \(=\frac{7}{3}+2=\frac{13}{3}\) square units
86856
The area of the curve bounded by \(x^{2}+y^{2}=4\) and \(x+y=4\) is
1 \(\pi\)
2 \(\pi-8\)
3 \(\pi+8\)
4 \(8-\pi\)
Explanation:
(D) : Area of curve bounded by \(x^{2}+y^{2}=4\) and \(x+y=4\) Now, area of \(\triangle \mathrm{AOB}\), \(\frac{1}{2} \times 4 \times 4=8\) Area of quarter circle \(\left(\frac{1}{4} \pi(2)^{2}\right)=\pi\) Area of shaded region \(=\) Area of triangle - Area of quarter circle \(=8-\pi\)
AMU-2004
Application of the Integrals
86864
Area bounded by the curve \(x=0\) and \(x+2|y|=\) 1 is :
86857
The area bounded by the curve \(y=2 x^{4}-x^{2}\), the \(x\)-axis and the two ordinates corresponding to minimal of the function is
1 \(\frac{1}{40}\) square unit
2 \(\frac{7}{120}\) square unit
3 \(\frac{1}{24}\) square unit
4 None of the above
Explanation:
(B): Given the curve, \(y=2 x^{4}-x^{2}\) \(\frac{d y}{d x}=8 x^{3}-2 x=0=4 x^{3}-x=0=x\left(4 x^{2}-1\right)=0\) \(\therefore \mathrm{x}=0, \quad 4 \mathrm{x}^{2}=1 \Rightarrow \mathrm{x}^{2}=\frac{1}{4}\) \(\mathrm{x}= \pm \frac{1}{2}\) Now, area of curves, \(A=\int_{-1 / 2}^{1 / 2}\left(2 x^{4}-x^{2}\right) d x\) \(A=2\left[\int_{0}^{1 / 2}\left(2 x^{4}-x^{2}\right) d x\right]\) (By using even property) \(=2\left[\frac{2 \mathrm{x}^{5}}{5}-\frac{\mathrm{x}^{3}}{3}\right]_{0}^{1 / 2}=2\left|2 \times \frac{1}{32 \times 5}-\frac{1}{8 \times 3}\right|\) \(=2 \times\left|\frac{1}{16 \times 5}-\frac{1}{8 \times 3}\right|\) \(=2\left(\frac{24-80}{16 \times 5 \times 8 \times 3}\right)=\frac{56 \times 2}{16 \times 5 \times 8 \times 3}=\frac{7}{120}\) square unit
SCRA-2014
Application of the Integrals
86858
The line \(x=\frac{\pi}{4}\) divides the area of the region bounded by \(y=\sin x, y=\cos x\) and \(x\)-axis \(\left(0 \leq x \leq \frac{\pi}{2}\right)\) into two regions of areas \(A_{1}\) and \(\mathbf{A}_{2}\). Then \(\mathbf{A}_{1}, \mathbf{A}_{2}\) equals
86859
The area bounded by \(y=x^{2}+2, x\)-axis, \(x=1\) and \(x=2\) is
1 \(\frac{16}{3}\) sq unit
2 \(\frac{17}{3}\) sq unit
3 \(\frac{13}{3}\) sq unit
4 \(\frac{20}{3}\) sq unit
Explanation:
(C) : Given curve, \(\mathrm{y}=\mathrm{x}^{2}+2\) \(\therefore\) The required area is \(A=\int_{1}^{2} y \cdot d x=\int_{1}^{2}\left(x^{2}+2\right) d x=\int_{1}^{2} x^{2} d x+\int_{1}^{2} 2 \cdot d x\) \(=\left[\frac{\mathrm{x}^{3}}{3}\right]_{1}^{2}+2 \cdot[\mathrm{x}]_{1}^{2}=\frac{1}{3}(8-1)+2(2-1)\) \(=\frac{7}{3}+2=\frac{13}{3}\) square units
86856
The area of the curve bounded by \(x^{2}+y^{2}=4\) and \(x+y=4\) is
1 \(\pi\)
2 \(\pi-8\)
3 \(\pi+8\)
4 \(8-\pi\)
Explanation:
(D) : Area of curve bounded by \(x^{2}+y^{2}=4\) and \(x+y=4\) Now, area of \(\triangle \mathrm{AOB}\), \(\frac{1}{2} \times 4 \times 4=8\) Area of quarter circle \(\left(\frac{1}{4} \pi(2)^{2}\right)=\pi\) Area of shaded region \(=\) Area of triangle - Area of quarter circle \(=8-\pi\)
AMU-2004
Application of the Integrals
86864
Area bounded by the curve \(x=0\) and \(x+2|y|=\) 1 is :
86857
The area bounded by the curve \(y=2 x^{4}-x^{2}\), the \(x\)-axis and the two ordinates corresponding to minimal of the function is
1 \(\frac{1}{40}\) square unit
2 \(\frac{7}{120}\) square unit
3 \(\frac{1}{24}\) square unit
4 None of the above
Explanation:
(B): Given the curve, \(y=2 x^{4}-x^{2}\) \(\frac{d y}{d x}=8 x^{3}-2 x=0=4 x^{3}-x=0=x\left(4 x^{2}-1\right)=0\) \(\therefore \mathrm{x}=0, \quad 4 \mathrm{x}^{2}=1 \Rightarrow \mathrm{x}^{2}=\frac{1}{4}\) \(\mathrm{x}= \pm \frac{1}{2}\) Now, area of curves, \(A=\int_{-1 / 2}^{1 / 2}\left(2 x^{4}-x^{2}\right) d x\) \(A=2\left[\int_{0}^{1 / 2}\left(2 x^{4}-x^{2}\right) d x\right]\) (By using even property) \(=2\left[\frac{2 \mathrm{x}^{5}}{5}-\frac{\mathrm{x}^{3}}{3}\right]_{0}^{1 / 2}=2\left|2 \times \frac{1}{32 \times 5}-\frac{1}{8 \times 3}\right|\) \(=2 \times\left|\frac{1}{16 \times 5}-\frac{1}{8 \times 3}\right|\) \(=2\left(\frac{24-80}{16 \times 5 \times 8 \times 3}\right)=\frac{56 \times 2}{16 \times 5 \times 8 \times 3}=\frac{7}{120}\) square unit
SCRA-2014
Application of the Integrals
86858
The line \(x=\frac{\pi}{4}\) divides the area of the region bounded by \(y=\sin x, y=\cos x\) and \(x\)-axis \(\left(0 \leq x \leq \frac{\pi}{2}\right)\) into two regions of areas \(A_{1}\) and \(\mathbf{A}_{2}\). Then \(\mathbf{A}_{1}, \mathbf{A}_{2}\) equals
86859
The area bounded by \(y=x^{2}+2, x\)-axis, \(x=1\) and \(x=2\) is
1 \(\frac{16}{3}\) sq unit
2 \(\frac{17}{3}\) sq unit
3 \(\frac{13}{3}\) sq unit
4 \(\frac{20}{3}\) sq unit
Explanation:
(C) : Given curve, \(\mathrm{y}=\mathrm{x}^{2}+2\) \(\therefore\) The required area is \(A=\int_{1}^{2} y \cdot d x=\int_{1}^{2}\left(x^{2}+2\right) d x=\int_{1}^{2} x^{2} d x+\int_{1}^{2} 2 \cdot d x\) \(=\left[\frac{\mathrm{x}^{3}}{3}\right]_{1}^{2}+2 \cdot[\mathrm{x}]_{1}^{2}=\frac{1}{3}(8-1)+2(2-1)\) \(=\frac{7}{3}+2=\frac{13}{3}\) square units
86856
The area of the curve bounded by \(x^{2}+y^{2}=4\) and \(x+y=4\) is
1 \(\pi\)
2 \(\pi-8\)
3 \(\pi+8\)
4 \(8-\pi\)
Explanation:
(D) : Area of curve bounded by \(x^{2}+y^{2}=4\) and \(x+y=4\) Now, area of \(\triangle \mathrm{AOB}\), \(\frac{1}{2} \times 4 \times 4=8\) Area of quarter circle \(\left(\frac{1}{4} \pi(2)^{2}\right)=\pi\) Area of shaded region \(=\) Area of triangle - Area of quarter circle \(=8-\pi\)
AMU-2004
Application of the Integrals
86864
Area bounded by the curve \(x=0\) and \(x+2|y|=\) 1 is :
86857
The area bounded by the curve \(y=2 x^{4}-x^{2}\), the \(x\)-axis and the two ordinates corresponding to minimal of the function is
1 \(\frac{1}{40}\) square unit
2 \(\frac{7}{120}\) square unit
3 \(\frac{1}{24}\) square unit
4 None of the above
Explanation:
(B): Given the curve, \(y=2 x^{4}-x^{2}\) \(\frac{d y}{d x}=8 x^{3}-2 x=0=4 x^{3}-x=0=x\left(4 x^{2}-1\right)=0\) \(\therefore \mathrm{x}=0, \quad 4 \mathrm{x}^{2}=1 \Rightarrow \mathrm{x}^{2}=\frac{1}{4}\) \(\mathrm{x}= \pm \frac{1}{2}\) Now, area of curves, \(A=\int_{-1 / 2}^{1 / 2}\left(2 x^{4}-x^{2}\right) d x\) \(A=2\left[\int_{0}^{1 / 2}\left(2 x^{4}-x^{2}\right) d x\right]\) (By using even property) \(=2\left[\frac{2 \mathrm{x}^{5}}{5}-\frac{\mathrm{x}^{3}}{3}\right]_{0}^{1 / 2}=2\left|2 \times \frac{1}{32 \times 5}-\frac{1}{8 \times 3}\right|\) \(=2 \times\left|\frac{1}{16 \times 5}-\frac{1}{8 \times 3}\right|\) \(=2\left(\frac{24-80}{16 \times 5 \times 8 \times 3}\right)=\frac{56 \times 2}{16 \times 5 \times 8 \times 3}=\frac{7}{120}\) square unit
SCRA-2014
Application of the Integrals
86858
The line \(x=\frac{\pi}{4}\) divides the area of the region bounded by \(y=\sin x, y=\cos x\) and \(x\)-axis \(\left(0 \leq x \leq \frac{\pi}{2}\right)\) into two regions of areas \(A_{1}\) and \(\mathbf{A}_{2}\). Then \(\mathbf{A}_{1}, \mathbf{A}_{2}\) equals
86859
The area bounded by \(y=x^{2}+2, x\)-axis, \(x=1\) and \(x=2\) is
1 \(\frac{16}{3}\) sq unit
2 \(\frac{17}{3}\) sq unit
3 \(\frac{13}{3}\) sq unit
4 \(\frac{20}{3}\) sq unit
Explanation:
(C) : Given curve, \(\mathrm{y}=\mathrm{x}^{2}+2\) \(\therefore\) The required area is \(A=\int_{1}^{2} y \cdot d x=\int_{1}^{2}\left(x^{2}+2\right) d x=\int_{1}^{2} x^{2} d x+\int_{1}^{2} 2 \cdot d x\) \(=\left[\frac{\mathrm{x}^{3}}{3}\right]_{1}^{2}+2 \cdot[\mathrm{x}]_{1}^{2}=\frac{1}{3}(8-1)+2(2-1)\) \(=\frac{7}{3}+2=\frac{13}{3}\) square units
86856
The area of the curve bounded by \(x^{2}+y^{2}=4\) and \(x+y=4\) is
1 \(\pi\)
2 \(\pi-8\)
3 \(\pi+8\)
4 \(8-\pi\)
Explanation:
(D) : Area of curve bounded by \(x^{2}+y^{2}=4\) and \(x+y=4\) Now, area of \(\triangle \mathrm{AOB}\), \(\frac{1}{2} \times 4 \times 4=8\) Area of quarter circle \(\left(\frac{1}{4} \pi(2)^{2}\right)=\pi\) Area of shaded region \(=\) Area of triangle - Area of quarter circle \(=8-\pi\)
AMU-2004
Application of the Integrals
86864
Area bounded by the curve \(x=0\) and \(x+2|y|=\) 1 is :
86857
The area bounded by the curve \(y=2 x^{4}-x^{2}\), the \(x\)-axis and the two ordinates corresponding to minimal of the function is
1 \(\frac{1}{40}\) square unit
2 \(\frac{7}{120}\) square unit
3 \(\frac{1}{24}\) square unit
4 None of the above
Explanation:
(B): Given the curve, \(y=2 x^{4}-x^{2}\) \(\frac{d y}{d x}=8 x^{3}-2 x=0=4 x^{3}-x=0=x\left(4 x^{2}-1\right)=0\) \(\therefore \mathrm{x}=0, \quad 4 \mathrm{x}^{2}=1 \Rightarrow \mathrm{x}^{2}=\frac{1}{4}\) \(\mathrm{x}= \pm \frac{1}{2}\) Now, area of curves, \(A=\int_{-1 / 2}^{1 / 2}\left(2 x^{4}-x^{2}\right) d x\) \(A=2\left[\int_{0}^{1 / 2}\left(2 x^{4}-x^{2}\right) d x\right]\) (By using even property) \(=2\left[\frac{2 \mathrm{x}^{5}}{5}-\frac{\mathrm{x}^{3}}{3}\right]_{0}^{1 / 2}=2\left|2 \times \frac{1}{32 \times 5}-\frac{1}{8 \times 3}\right|\) \(=2 \times\left|\frac{1}{16 \times 5}-\frac{1}{8 \times 3}\right|\) \(=2\left(\frac{24-80}{16 \times 5 \times 8 \times 3}\right)=\frac{56 \times 2}{16 \times 5 \times 8 \times 3}=\frac{7}{120}\) square unit
SCRA-2014
Application of the Integrals
86858
The line \(x=\frac{\pi}{4}\) divides the area of the region bounded by \(y=\sin x, y=\cos x\) and \(x\)-axis \(\left(0 \leq x \leq \frac{\pi}{2}\right)\) into two regions of areas \(A_{1}\) and \(\mathbf{A}_{2}\). Then \(\mathbf{A}_{1}, \mathbf{A}_{2}\) equals
86859
The area bounded by \(y=x^{2}+2, x\)-axis, \(x=1\) and \(x=2\) is
1 \(\frac{16}{3}\) sq unit
2 \(\frac{17}{3}\) sq unit
3 \(\frac{13}{3}\) sq unit
4 \(\frac{20}{3}\) sq unit
Explanation:
(C) : Given curve, \(\mathrm{y}=\mathrm{x}^{2}+2\) \(\therefore\) The required area is \(A=\int_{1}^{2} y \cdot d x=\int_{1}^{2}\left(x^{2}+2\right) d x=\int_{1}^{2} x^{2} d x+\int_{1}^{2} 2 \cdot d x\) \(=\left[\frac{\mathrm{x}^{3}}{3}\right]_{1}^{2}+2 \cdot[\mathrm{x}]_{1}^{2}=\frac{1}{3}(8-1)+2(2-1)\) \(=\frac{7}{3}+2=\frac{13}{3}\) square units
