86963
The area bounded by \(y=x^{2}+3\) and \(y=2 x+3\) is (in sq units)
1 \(\frac{12}{7}\)
2 \(\frac{4}{3}\)
3 \(\frac{3}{4}\)
4 \(\frac{8}{3}\)
5 \(\frac{3}{8}\)
Explanation:
(B) : Given, The area bounded by \(y=x^{2}+3\) and \(y=2 x+3\) Area, \(\quad(A)=\int_{0}^{2}(2 x+3)-\left(x^{2}+3\right) d x\) \(=\int_{0}^{2}\left(2 x-x^{2}\right) d x=\left[\frac{2 x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{2}=\left[4-\frac{8}{3}\right]\) \(A=\frac{4}{3}\) square units
Kerala CEE-2015
Application of the Integrals
86873
The area bounded by the curve \(y=x^{2}\), the \(y\) axis and the straight lines \(y=1\) and \(y=4\) is
1 \(\frac{14}{3}\) sq.units
2 \(\frac{11}{3}\) sq.units
3 \(\frac{15}{5}\) sq.units
4 \(\frac{1}{3}\) sq.units
Explanation:
(A) : Given, Curve, \(y=x^{2}\) \(x=\sqrt{y}=y^{\frac{1}{2}}\) Limits from, \(\mathrm{y}=1\) and \(\mathrm{y}=4\) Required area \(=\int_{1}^{4} \sqrt{y} d y=\int_{1}^{4}(y)^{\frac{1}{2}} d y\) \({\left[\frac{\left.\mathrm{y}^{\left(\frac{1}{2}+1\right.}\right)}{\frac{1}{2}+1}\right]_{1}^{4} } =\left[\frac{\mathrm{y}^{\frac{3}{2}}}{\frac{3}{2}}\right]_{1}^{4}=\frac{2}{3}(4)^{\frac{3}{2}}-\frac{2}{3}(\mathrm{y})^{\frac{3}{2}}=\frac{2}{3} \times 8-\frac{2}{3} \times 1\) \(=\frac{16}{3}-\frac{2}{3}=\frac{14}{3} \text { Square units }\)
J&K CET-2016
Application of the Integrals
86874
The area enclosed between the curves \(y^{2}=x\) and \(\mathbf{y}=|\mathbf{x}|\) is
1 \(\frac{2}{3}\) squnit
2 1squnit
3 \(\frac{1}{6}\) squnit
4 \(\frac{1}{3}\) squnit
Explanation:
(C) : Point of intersection \(-y^{2}=x\) and \(y=|x|\) Or \(y^{2}=x\) \(x^{2}=x\) Intersecting point, \(\mathrm{x}=1\) and \(\mathrm{x}=0\) Area of enclosed \(=\int_{0}^{1} \sqrt{x}-x\) \(=\left[\frac{2}{3} x^{3 / 2}-\frac{x^{2}}{2}\right]_{0}^{1}=\frac{2}{3}-\frac{1}{2} \Rightarrow=\frac{4-3}{6}=\frac{1}{6}\) square units.
AIEEE-2007
Application of the Integrals
86875
The area (in sq units) of the region bounded by the parabola, \(y=x^{2}+2\) and the lines, \(y=x+1\), \(x=0\) and \(x=3\) is
1 \(\frac{17}{4}\)
2 \(\frac{21}{2}\)
3 \(\frac{15}{4}\)
4 \(\frac{15}{2}\)
Explanation:
(A) : Given, parabola \(\therefore \quad \mathrm{y}=\mathrm{x}^{2}+2\) and the line \(\mathrm{y}=\mathrm{x}+1\) Required area \(=\int_{0}^{3}\left(x^{2}+2\right) d x-\int_{0}^{3}(x+1) d x\) \(=\left[\frac{x^{3}}{3}+2 x\right]_{0}^{3}-\left[\frac{x^{2}}{2}+x\right]_{0}^{3}=\frac{27}{3}+6-\frac{9}{2}+3\) \(=15-\frac{9}{2}-3=\frac{30-9-6}{2}=\frac{15}{2}\)
Application of the Integrals
86876
The area (in sq units) of the region bounded by the curves \(y=2^{x}\) and \(y=|x+1|\) in the first quadrants is
1 \(\frac{3}{2}\)
2 \(\log _{\mathrm{e}} 2+\frac{3}{2}\)
3 \(\frac{1}{2}\)
4 \(\frac{3}{2}-\frac{1}{\log _{\mathrm{e}} 2}\)
Explanation:
(D) : Given, \(\mathrm{y}=2^{\mathrm{x}} \quad \text { and } \mathrm{y}=|\mathrm{x}+1|\) Required area \(=\int_{0}^{1}\left((x+1)-2^{x}\right) d x\) \(=\left(\frac{x^{2}}{2}+x-\frac{2^{x}}{\log _{e} 2}\right)_{0}^{1}=\left(\frac{1}{2}+1-\frac{2}{\log _{e} 2}+\frac{1}{\log _{e} 2}\right)\) \(=\left(\frac{3}{2}-\frac{1}{\log _{\mathrm{e}} 2}\right)\)
86963
The area bounded by \(y=x^{2}+3\) and \(y=2 x+3\) is (in sq units)
1 \(\frac{12}{7}\)
2 \(\frac{4}{3}\)
3 \(\frac{3}{4}\)
4 \(\frac{8}{3}\)
5 \(\frac{3}{8}\)
Explanation:
(B) : Given, The area bounded by \(y=x^{2}+3\) and \(y=2 x+3\) Area, \(\quad(A)=\int_{0}^{2}(2 x+3)-\left(x^{2}+3\right) d x\) \(=\int_{0}^{2}\left(2 x-x^{2}\right) d x=\left[\frac{2 x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{2}=\left[4-\frac{8}{3}\right]\) \(A=\frac{4}{3}\) square units
Kerala CEE-2015
Application of the Integrals
86873
The area bounded by the curve \(y=x^{2}\), the \(y\) axis and the straight lines \(y=1\) and \(y=4\) is
1 \(\frac{14}{3}\) sq.units
2 \(\frac{11}{3}\) sq.units
3 \(\frac{15}{5}\) sq.units
4 \(\frac{1}{3}\) sq.units
Explanation:
(A) : Given, Curve, \(y=x^{2}\) \(x=\sqrt{y}=y^{\frac{1}{2}}\) Limits from, \(\mathrm{y}=1\) and \(\mathrm{y}=4\) Required area \(=\int_{1}^{4} \sqrt{y} d y=\int_{1}^{4}(y)^{\frac{1}{2}} d y\) \({\left[\frac{\left.\mathrm{y}^{\left(\frac{1}{2}+1\right.}\right)}{\frac{1}{2}+1}\right]_{1}^{4} } =\left[\frac{\mathrm{y}^{\frac{3}{2}}}{\frac{3}{2}}\right]_{1}^{4}=\frac{2}{3}(4)^{\frac{3}{2}}-\frac{2}{3}(\mathrm{y})^{\frac{3}{2}}=\frac{2}{3} \times 8-\frac{2}{3} \times 1\) \(=\frac{16}{3}-\frac{2}{3}=\frac{14}{3} \text { Square units }\)
J&K CET-2016
Application of the Integrals
86874
The area enclosed between the curves \(y^{2}=x\) and \(\mathbf{y}=|\mathbf{x}|\) is
1 \(\frac{2}{3}\) squnit
2 1squnit
3 \(\frac{1}{6}\) squnit
4 \(\frac{1}{3}\) squnit
Explanation:
(C) : Point of intersection \(-y^{2}=x\) and \(y=|x|\) Or \(y^{2}=x\) \(x^{2}=x\) Intersecting point, \(\mathrm{x}=1\) and \(\mathrm{x}=0\) Area of enclosed \(=\int_{0}^{1} \sqrt{x}-x\) \(=\left[\frac{2}{3} x^{3 / 2}-\frac{x^{2}}{2}\right]_{0}^{1}=\frac{2}{3}-\frac{1}{2} \Rightarrow=\frac{4-3}{6}=\frac{1}{6}\) square units.
AIEEE-2007
Application of the Integrals
86875
The area (in sq units) of the region bounded by the parabola, \(y=x^{2}+2\) and the lines, \(y=x+1\), \(x=0\) and \(x=3\) is
1 \(\frac{17}{4}\)
2 \(\frac{21}{2}\)
3 \(\frac{15}{4}\)
4 \(\frac{15}{2}\)
Explanation:
(A) : Given, parabola \(\therefore \quad \mathrm{y}=\mathrm{x}^{2}+2\) and the line \(\mathrm{y}=\mathrm{x}+1\) Required area \(=\int_{0}^{3}\left(x^{2}+2\right) d x-\int_{0}^{3}(x+1) d x\) \(=\left[\frac{x^{3}}{3}+2 x\right]_{0}^{3}-\left[\frac{x^{2}}{2}+x\right]_{0}^{3}=\frac{27}{3}+6-\frac{9}{2}+3\) \(=15-\frac{9}{2}-3=\frac{30-9-6}{2}=\frac{15}{2}\)
Application of the Integrals
86876
The area (in sq units) of the region bounded by the curves \(y=2^{x}\) and \(y=|x+1|\) in the first quadrants is
1 \(\frac{3}{2}\)
2 \(\log _{\mathrm{e}} 2+\frac{3}{2}\)
3 \(\frac{1}{2}\)
4 \(\frac{3}{2}-\frac{1}{\log _{\mathrm{e}} 2}\)
Explanation:
(D) : Given, \(\mathrm{y}=2^{\mathrm{x}} \quad \text { and } \mathrm{y}=|\mathrm{x}+1|\) Required area \(=\int_{0}^{1}\left((x+1)-2^{x}\right) d x\) \(=\left(\frac{x^{2}}{2}+x-\frac{2^{x}}{\log _{e} 2}\right)_{0}^{1}=\left(\frac{1}{2}+1-\frac{2}{\log _{e} 2}+\frac{1}{\log _{e} 2}\right)\) \(=\left(\frac{3}{2}-\frac{1}{\log _{\mathrm{e}} 2}\right)\)
86963
The area bounded by \(y=x^{2}+3\) and \(y=2 x+3\) is (in sq units)
1 \(\frac{12}{7}\)
2 \(\frac{4}{3}\)
3 \(\frac{3}{4}\)
4 \(\frac{8}{3}\)
5 \(\frac{3}{8}\)
Explanation:
(B) : Given, The area bounded by \(y=x^{2}+3\) and \(y=2 x+3\) Area, \(\quad(A)=\int_{0}^{2}(2 x+3)-\left(x^{2}+3\right) d x\) \(=\int_{0}^{2}\left(2 x-x^{2}\right) d x=\left[\frac{2 x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{2}=\left[4-\frac{8}{3}\right]\) \(A=\frac{4}{3}\) square units
Kerala CEE-2015
Application of the Integrals
86873
The area bounded by the curve \(y=x^{2}\), the \(y\) axis and the straight lines \(y=1\) and \(y=4\) is
1 \(\frac{14}{3}\) sq.units
2 \(\frac{11}{3}\) sq.units
3 \(\frac{15}{5}\) sq.units
4 \(\frac{1}{3}\) sq.units
Explanation:
(A) : Given, Curve, \(y=x^{2}\) \(x=\sqrt{y}=y^{\frac{1}{2}}\) Limits from, \(\mathrm{y}=1\) and \(\mathrm{y}=4\) Required area \(=\int_{1}^{4} \sqrt{y} d y=\int_{1}^{4}(y)^{\frac{1}{2}} d y\) \({\left[\frac{\left.\mathrm{y}^{\left(\frac{1}{2}+1\right.}\right)}{\frac{1}{2}+1}\right]_{1}^{4} } =\left[\frac{\mathrm{y}^{\frac{3}{2}}}{\frac{3}{2}}\right]_{1}^{4}=\frac{2}{3}(4)^{\frac{3}{2}}-\frac{2}{3}(\mathrm{y})^{\frac{3}{2}}=\frac{2}{3} \times 8-\frac{2}{3} \times 1\) \(=\frac{16}{3}-\frac{2}{3}=\frac{14}{3} \text { Square units }\)
J&K CET-2016
Application of the Integrals
86874
The area enclosed between the curves \(y^{2}=x\) and \(\mathbf{y}=|\mathbf{x}|\) is
1 \(\frac{2}{3}\) squnit
2 1squnit
3 \(\frac{1}{6}\) squnit
4 \(\frac{1}{3}\) squnit
Explanation:
(C) : Point of intersection \(-y^{2}=x\) and \(y=|x|\) Or \(y^{2}=x\) \(x^{2}=x\) Intersecting point, \(\mathrm{x}=1\) and \(\mathrm{x}=0\) Area of enclosed \(=\int_{0}^{1} \sqrt{x}-x\) \(=\left[\frac{2}{3} x^{3 / 2}-\frac{x^{2}}{2}\right]_{0}^{1}=\frac{2}{3}-\frac{1}{2} \Rightarrow=\frac{4-3}{6}=\frac{1}{6}\) square units.
AIEEE-2007
Application of the Integrals
86875
The area (in sq units) of the region bounded by the parabola, \(y=x^{2}+2\) and the lines, \(y=x+1\), \(x=0\) and \(x=3\) is
1 \(\frac{17}{4}\)
2 \(\frac{21}{2}\)
3 \(\frac{15}{4}\)
4 \(\frac{15}{2}\)
Explanation:
(A) : Given, parabola \(\therefore \quad \mathrm{y}=\mathrm{x}^{2}+2\) and the line \(\mathrm{y}=\mathrm{x}+1\) Required area \(=\int_{0}^{3}\left(x^{2}+2\right) d x-\int_{0}^{3}(x+1) d x\) \(=\left[\frac{x^{3}}{3}+2 x\right]_{0}^{3}-\left[\frac{x^{2}}{2}+x\right]_{0}^{3}=\frac{27}{3}+6-\frac{9}{2}+3\) \(=15-\frac{9}{2}-3=\frac{30-9-6}{2}=\frac{15}{2}\)
Application of the Integrals
86876
The area (in sq units) of the region bounded by the curves \(y=2^{x}\) and \(y=|x+1|\) in the first quadrants is
1 \(\frac{3}{2}\)
2 \(\log _{\mathrm{e}} 2+\frac{3}{2}\)
3 \(\frac{1}{2}\)
4 \(\frac{3}{2}-\frac{1}{\log _{\mathrm{e}} 2}\)
Explanation:
(D) : Given, \(\mathrm{y}=2^{\mathrm{x}} \quad \text { and } \mathrm{y}=|\mathrm{x}+1|\) Required area \(=\int_{0}^{1}\left((x+1)-2^{x}\right) d x\) \(=\left(\frac{x^{2}}{2}+x-\frac{2^{x}}{\log _{e} 2}\right)_{0}^{1}=\left(\frac{1}{2}+1-\frac{2}{\log _{e} 2}+\frac{1}{\log _{e} 2}\right)\) \(=\left(\frac{3}{2}-\frac{1}{\log _{\mathrm{e}} 2}\right)\)
86963
The area bounded by \(y=x^{2}+3\) and \(y=2 x+3\) is (in sq units)
1 \(\frac{12}{7}\)
2 \(\frac{4}{3}\)
3 \(\frac{3}{4}\)
4 \(\frac{8}{3}\)
5 \(\frac{3}{8}\)
Explanation:
(B) : Given, The area bounded by \(y=x^{2}+3\) and \(y=2 x+3\) Area, \(\quad(A)=\int_{0}^{2}(2 x+3)-\left(x^{2}+3\right) d x\) \(=\int_{0}^{2}\left(2 x-x^{2}\right) d x=\left[\frac{2 x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{2}=\left[4-\frac{8}{3}\right]\) \(A=\frac{4}{3}\) square units
Kerala CEE-2015
Application of the Integrals
86873
The area bounded by the curve \(y=x^{2}\), the \(y\) axis and the straight lines \(y=1\) and \(y=4\) is
1 \(\frac{14}{3}\) sq.units
2 \(\frac{11}{3}\) sq.units
3 \(\frac{15}{5}\) sq.units
4 \(\frac{1}{3}\) sq.units
Explanation:
(A) : Given, Curve, \(y=x^{2}\) \(x=\sqrt{y}=y^{\frac{1}{2}}\) Limits from, \(\mathrm{y}=1\) and \(\mathrm{y}=4\) Required area \(=\int_{1}^{4} \sqrt{y} d y=\int_{1}^{4}(y)^{\frac{1}{2}} d y\) \({\left[\frac{\left.\mathrm{y}^{\left(\frac{1}{2}+1\right.}\right)}{\frac{1}{2}+1}\right]_{1}^{4} } =\left[\frac{\mathrm{y}^{\frac{3}{2}}}{\frac{3}{2}}\right]_{1}^{4}=\frac{2}{3}(4)^{\frac{3}{2}}-\frac{2}{3}(\mathrm{y})^{\frac{3}{2}}=\frac{2}{3} \times 8-\frac{2}{3} \times 1\) \(=\frac{16}{3}-\frac{2}{3}=\frac{14}{3} \text { Square units }\)
J&K CET-2016
Application of the Integrals
86874
The area enclosed between the curves \(y^{2}=x\) and \(\mathbf{y}=|\mathbf{x}|\) is
1 \(\frac{2}{3}\) squnit
2 1squnit
3 \(\frac{1}{6}\) squnit
4 \(\frac{1}{3}\) squnit
Explanation:
(C) : Point of intersection \(-y^{2}=x\) and \(y=|x|\) Or \(y^{2}=x\) \(x^{2}=x\) Intersecting point, \(\mathrm{x}=1\) and \(\mathrm{x}=0\) Area of enclosed \(=\int_{0}^{1} \sqrt{x}-x\) \(=\left[\frac{2}{3} x^{3 / 2}-\frac{x^{2}}{2}\right]_{0}^{1}=\frac{2}{3}-\frac{1}{2} \Rightarrow=\frac{4-3}{6}=\frac{1}{6}\) square units.
AIEEE-2007
Application of the Integrals
86875
The area (in sq units) of the region bounded by the parabola, \(y=x^{2}+2\) and the lines, \(y=x+1\), \(x=0\) and \(x=3\) is
1 \(\frac{17}{4}\)
2 \(\frac{21}{2}\)
3 \(\frac{15}{4}\)
4 \(\frac{15}{2}\)
Explanation:
(A) : Given, parabola \(\therefore \quad \mathrm{y}=\mathrm{x}^{2}+2\) and the line \(\mathrm{y}=\mathrm{x}+1\) Required area \(=\int_{0}^{3}\left(x^{2}+2\right) d x-\int_{0}^{3}(x+1) d x\) \(=\left[\frac{x^{3}}{3}+2 x\right]_{0}^{3}-\left[\frac{x^{2}}{2}+x\right]_{0}^{3}=\frac{27}{3}+6-\frac{9}{2}+3\) \(=15-\frac{9}{2}-3=\frac{30-9-6}{2}=\frac{15}{2}\)
Application of the Integrals
86876
The area (in sq units) of the region bounded by the curves \(y=2^{x}\) and \(y=|x+1|\) in the first quadrants is
1 \(\frac{3}{2}\)
2 \(\log _{\mathrm{e}} 2+\frac{3}{2}\)
3 \(\frac{1}{2}\)
4 \(\frac{3}{2}-\frac{1}{\log _{\mathrm{e}} 2}\)
Explanation:
(D) : Given, \(\mathrm{y}=2^{\mathrm{x}} \quad \text { and } \mathrm{y}=|\mathrm{x}+1|\) Required area \(=\int_{0}^{1}\left((x+1)-2^{x}\right) d x\) \(=\left(\frac{x^{2}}{2}+x-\frac{2^{x}}{\log _{e} 2}\right)_{0}^{1}=\left(\frac{1}{2}+1-\frac{2}{\log _{e} 2}+\frac{1}{\log _{e} 2}\right)\) \(=\left(\frac{3}{2}-\frac{1}{\log _{\mathrm{e}} 2}\right)\)
86963
The area bounded by \(y=x^{2}+3\) and \(y=2 x+3\) is (in sq units)
1 \(\frac{12}{7}\)
2 \(\frac{4}{3}\)
3 \(\frac{3}{4}\)
4 \(\frac{8}{3}\)
5 \(\frac{3}{8}\)
Explanation:
(B) : Given, The area bounded by \(y=x^{2}+3\) and \(y=2 x+3\) Area, \(\quad(A)=\int_{0}^{2}(2 x+3)-\left(x^{2}+3\right) d x\) \(=\int_{0}^{2}\left(2 x-x^{2}\right) d x=\left[\frac{2 x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{2}=\left[4-\frac{8}{3}\right]\) \(A=\frac{4}{3}\) square units
Kerala CEE-2015
Application of the Integrals
86873
The area bounded by the curve \(y=x^{2}\), the \(y\) axis and the straight lines \(y=1\) and \(y=4\) is
1 \(\frac{14}{3}\) sq.units
2 \(\frac{11}{3}\) sq.units
3 \(\frac{15}{5}\) sq.units
4 \(\frac{1}{3}\) sq.units
Explanation:
(A) : Given, Curve, \(y=x^{2}\) \(x=\sqrt{y}=y^{\frac{1}{2}}\) Limits from, \(\mathrm{y}=1\) and \(\mathrm{y}=4\) Required area \(=\int_{1}^{4} \sqrt{y} d y=\int_{1}^{4}(y)^{\frac{1}{2}} d y\) \({\left[\frac{\left.\mathrm{y}^{\left(\frac{1}{2}+1\right.}\right)}{\frac{1}{2}+1}\right]_{1}^{4} } =\left[\frac{\mathrm{y}^{\frac{3}{2}}}{\frac{3}{2}}\right]_{1}^{4}=\frac{2}{3}(4)^{\frac{3}{2}}-\frac{2}{3}(\mathrm{y})^{\frac{3}{2}}=\frac{2}{3} \times 8-\frac{2}{3} \times 1\) \(=\frac{16}{3}-\frac{2}{3}=\frac{14}{3} \text { Square units }\)
J&K CET-2016
Application of the Integrals
86874
The area enclosed between the curves \(y^{2}=x\) and \(\mathbf{y}=|\mathbf{x}|\) is
1 \(\frac{2}{3}\) squnit
2 1squnit
3 \(\frac{1}{6}\) squnit
4 \(\frac{1}{3}\) squnit
Explanation:
(C) : Point of intersection \(-y^{2}=x\) and \(y=|x|\) Or \(y^{2}=x\) \(x^{2}=x\) Intersecting point, \(\mathrm{x}=1\) and \(\mathrm{x}=0\) Area of enclosed \(=\int_{0}^{1} \sqrt{x}-x\) \(=\left[\frac{2}{3} x^{3 / 2}-\frac{x^{2}}{2}\right]_{0}^{1}=\frac{2}{3}-\frac{1}{2} \Rightarrow=\frac{4-3}{6}=\frac{1}{6}\) square units.
AIEEE-2007
Application of the Integrals
86875
The area (in sq units) of the region bounded by the parabola, \(y=x^{2}+2\) and the lines, \(y=x+1\), \(x=0\) and \(x=3\) is
1 \(\frac{17}{4}\)
2 \(\frac{21}{2}\)
3 \(\frac{15}{4}\)
4 \(\frac{15}{2}\)
Explanation:
(A) : Given, parabola \(\therefore \quad \mathrm{y}=\mathrm{x}^{2}+2\) and the line \(\mathrm{y}=\mathrm{x}+1\) Required area \(=\int_{0}^{3}\left(x^{2}+2\right) d x-\int_{0}^{3}(x+1) d x\) \(=\left[\frac{x^{3}}{3}+2 x\right]_{0}^{3}-\left[\frac{x^{2}}{2}+x\right]_{0}^{3}=\frac{27}{3}+6-\frac{9}{2}+3\) \(=15-\frac{9}{2}-3=\frac{30-9-6}{2}=\frac{15}{2}\)
Application of the Integrals
86876
The area (in sq units) of the region bounded by the curves \(y=2^{x}\) and \(y=|x+1|\) in the first quadrants is
1 \(\frac{3}{2}\)
2 \(\log _{\mathrm{e}} 2+\frac{3}{2}\)
3 \(\frac{1}{2}\)
4 \(\frac{3}{2}-\frac{1}{\log _{\mathrm{e}} 2}\)
Explanation:
(D) : Given, \(\mathrm{y}=2^{\mathrm{x}} \quad \text { and } \mathrm{y}=|\mathrm{x}+1|\) Required area \(=\int_{0}^{1}\left((x+1)-2^{x}\right) d x\) \(=\left(\frac{x^{2}}{2}+x-\frac{2^{x}}{\log _{e} 2}\right)_{0}^{1}=\left(\frac{1}{2}+1-\frac{2}{\log _{e} 2}+\frac{1}{\log _{e} 2}\right)\) \(=\left(\frac{3}{2}-\frac{1}{\log _{\mathrm{e}} 2}\right)\)
