86598
\(\int_{-a}^{a} f(x) d x-\int_{0}^{a} f(-x) d x=\)
1 \(\int_{-a}^{a} f(a-x) d x\)
2 \(\int_{-a}^{a} f(x)+f(a-x) d x\)
3 \(\int_{0}^{a} f(x)+f(a-x) d x\)
4 \(\int_{0}^{a} f(a-x) d x\)
Explanation:
(D) : Let, \(I=\int_{-a}^{a} f(x) d x-\int_{0}^{a} f(-x) d x\) If \(f(x)\) is an even function, Then, \(f(-x)=f(x)\) \(I=2 \int_{0}^{a} f(-x) d x-\int_{0}^{a} f(-x) d x\) \(I=\int_{0}^{a} f(-x) d x=\int_{0}^{a} f(x) d x\) \(\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\) If \(f(x)\) is an odd function, Then, \(f(-x)=-f(x)\) And, \(\int_{-a}^{a} f(x) d x=0\) \(I=0+\int_{0}^{a} f(x) d x\) \(I=\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\)
AP EAMCET-2022-06.07.2022
Integral Calculus
86599
If \(\int_{0}^{1} f(x) d x=1, \int_{0}^{1} x f(x) d x=a, \int_{0}^{1} x^{2} f(x) d x=a^{2}\) then \(\int_{0}^{1}(x-a)^{2} f(x) d x=\)
1 \(a^{2}\)
2 \(\mathrm{a}^{2}+1\)
3 \(a^{2}-1\)
4 0
Explanation:
(D)Let, \(\left.I=\int_{0}^{1}(a-x)^{2}\right) f(x) d x\) \(I=\int_{0}^{1}\left(a^{2}+x^{2}-2 a x\right) f(x) d x\) \(I=\int_{0}^{1} a^{2} f(x) d x+\int_{0}^{1} x^{2} f(x) d x-\int_{0}^{1} 2 a x f(x) d x\) \(I=a^{2} \int_{0}^{1} f(x) d x+\int_{0}^{1} x^{2} f(x) d x-2 a \int_{0}^{1} x f(x) d x\) Given, \(a^{2} \int_{0}^{1} f(x) d x=1, \int_{0}^{1} x f(x) d x=a, \int_{0}^{1} x^{2} f(x) d=a^{2}\) \(I \stackrel{0}{=} a^{2} \times 1+a^{2}-2 a \times a\) \(I=2 a^{2}-2 a^{2}\) \(\mathrm{I}=0\)
AP EAMCET-2021-23.08.2021
Integral Calculus
86609
If \(I=\int_{0}^{1} \frac{\sin x}{\sqrt{x}} d x\) and \(J=\int_{0}^{1} \frac{\cos x}{\sqrt{x}} d x\). Then , which one of the following is true?
86598
\(\int_{-a}^{a} f(x) d x-\int_{0}^{a} f(-x) d x=\)
1 \(\int_{-a}^{a} f(a-x) d x\)
2 \(\int_{-a}^{a} f(x)+f(a-x) d x\)
3 \(\int_{0}^{a} f(x)+f(a-x) d x\)
4 \(\int_{0}^{a} f(a-x) d x\)
Explanation:
(D) : Let, \(I=\int_{-a}^{a} f(x) d x-\int_{0}^{a} f(-x) d x\) If \(f(x)\) is an even function, Then, \(f(-x)=f(x)\) \(I=2 \int_{0}^{a} f(-x) d x-\int_{0}^{a} f(-x) d x\) \(I=\int_{0}^{a} f(-x) d x=\int_{0}^{a} f(x) d x\) \(\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\) If \(f(x)\) is an odd function, Then, \(f(-x)=-f(x)\) And, \(\int_{-a}^{a} f(x) d x=0\) \(I=0+\int_{0}^{a} f(x) d x\) \(I=\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\)
AP EAMCET-2022-06.07.2022
Integral Calculus
86599
If \(\int_{0}^{1} f(x) d x=1, \int_{0}^{1} x f(x) d x=a, \int_{0}^{1} x^{2} f(x) d x=a^{2}\) then \(\int_{0}^{1}(x-a)^{2} f(x) d x=\)
1 \(a^{2}\)
2 \(\mathrm{a}^{2}+1\)
3 \(a^{2}-1\)
4 0
Explanation:
(D)Let, \(\left.I=\int_{0}^{1}(a-x)^{2}\right) f(x) d x\) \(I=\int_{0}^{1}\left(a^{2}+x^{2}-2 a x\right) f(x) d x\) \(I=\int_{0}^{1} a^{2} f(x) d x+\int_{0}^{1} x^{2} f(x) d x-\int_{0}^{1} 2 a x f(x) d x\) \(I=a^{2} \int_{0}^{1} f(x) d x+\int_{0}^{1} x^{2} f(x) d x-2 a \int_{0}^{1} x f(x) d x\) Given, \(a^{2} \int_{0}^{1} f(x) d x=1, \int_{0}^{1} x f(x) d x=a, \int_{0}^{1} x^{2} f(x) d=a^{2}\) \(I \stackrel{0}{=} a^{2} \times 1+a^{2}-2 a \times a\) \(I=2 a^{2}-2 a^{2}\) \(\mathrm{I}=0\)
AP EAMCET-2021-23.08.2021
Integral Calculus
86609
If \(I=\int_{0}^{1} \frac{\sin x}{\sqrt{x}} d x\) and \(J=\int_{0}^{1} \frac{\cos x}{\sqrt{x}} d x\). Then , which one of the following is true?
86598
\(\int_{-a}^{a} f(x) d x-\int_{0}^{a} f(-x) d x=\)
1 \(\int_{-a}^{a} f(a-x) d x\)
2 \(\int_{-a}^{a} f(x)+f(a-x) d x\)
3 \(\int_{0}^{a} f(x)+f(a-x) d x\)
4 \(\int_{0}^{a} f(a-x) d x\)
Explanation:
(D) : Let, \(I=\int_{-a}^{a} f(x) d x-\int_{0}^{a} f(-x) d x\) If \(f(x)\) is an even function, Then, \(f(-x)=f(x)\) \(I=2 \int_{0}^{a} f(-x) d x-\int_{0}^{a} f(-x) d x\) \(I=\int_{0}^{a} f(-x) d x=\int_{0}^{a} f(x) d x\) \(\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\) If \(f(x)\) is an odd function, Then, \(f(-x)=-f(x)\) And, \(\int_{-a}^{a} f(x) d x=0\) \(I=0+\int_{0}^{a} f(x) d x\) \(I=\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\)
AP EAMCET-2022-06.07.2022
Integral Calculus
86599
If \(\int_{0}^{1} f(x) d x=1, \int_{0}^{1} x f(x) d x=a, \int_{0}^{1} x^{2} f(x) d x=a^{2}\) then \(\int_{0}^{1}(x-a)^{2} f(x) d x=\)
1 \(a^{2}\)
2 \(\mathrm{a}^{2}+1\)
3 \(a^{2}-1\)
4 0
Explanation:
(D)Let, \(\left.I=\int_{0}^{1}(a-x)^{2}\right) f(x) d x\) \(I=\int_{0}^{1}\left(a^{2}+x^{2}-2 a x\right) f(x) d x\) \(I=\int_{0}^{1} a^{2} f(x) d x+\int_{0}^{1} x^{2} f(x) d x-\int_{0}^{1} 2 a x f(x) d x\) \(I=a^{2} \int_{0}^{1} f(x) d x+\int_{0}^{1} x^{2} f(x) d x-2 a \int_{0}^{1} x f(x) d x\) Given, \(a^{2} \int_{0}^{1} f(x) d x=1, \int_{0}^{1} x f(x) d x=a, \int_{0}^{1} x^{2} f(x) d=a^{2}\) \(I \stackrel{0}{=} a^{2} \times 1+a^{2}-2 a \times a\) \(I=2 a^{2}-2 a^{2}\) \(\mathrm{I}=0\)
AP EAMCET-2021-23.08.2021
Integral Calculus
86609
If \(I=\int_{0}^{1} \frac{\sin x}{\sqrt{x}} d x\) and \(J=\int_{0}^{1} \frac{\cos x}{\sqrt{x}} d x\). Then , which one of the following is true?
86598
\(\int_{-a}^{a} f(x) d x-\int_{0}^{a} f(-x) d x=\)
1 \(\int_{-a}^{a} f(a-x) d x\)
2 \(\int_{-a}^{a} f(x)+f(a-x) d x\)
3 \(\int_{0}^{a} f(x)+f(a-x) d x\)
4 \(\int_{0}^{a} f(a-x) d x\)
Explanation:
(D) : Let, \(I=\int_{-a}^{a} f(x) d x-\int_{0}^{a} f(-x) d x\) If \(f(x)\) is an even function, Then, \(f(-x)=f(x)\) \(I=2 \int_{0}^{a} f(-x) d x-\int_{0}^{a} f(-x) d x\) \(I=\int_{0}^{a} f(-x) d x=\int_{0}^{a} f(x) d x\) \(\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\) If \(f(x)\) is an odd function, Then, \(f(-x)=-f(x)\) And, \(\int_{-a}^{a} f(x) d x=0\) \(I=0+\int_{0}^{a} f(x) d x\) \(I=\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\)
AP EAMCET-2022-06.07.2022
Integral Calculus
86599
If \(\int_{0}^{1} f(x) d x=1, \int_{0}^{1} x f(x) d x=a, \int_{0}^{1} x^{2} f(x) d x=a^{2}\) then \(\int_{0}^{1}(x-a)^{2} f(x) d x=\)
1 \(a^{2}\)
2 \(\mathrm{a}^{2}+1\)
3 \(a^{2}-1\)
4 0
Explanation:
(D)Let, \(\left.I=\int_{0}^{1}(a-x)^{2}\right) f(x) d x\) \(I=\int_{0}^{1}\left(a^{2}+x^{2}-2 a x\right) f(x) d x\) \(I=\int_{0}^{1} a^{2} f(x) d x+\int_{0}^{1} x^{2} f(x) d x-\int_{0}^{1} 2 a x f(x) d x\) \(I=a^{2} \int_{0}^{1} f(x) d x+\int_{0}^{1} x^{2} f(x) d x-2 a \int_{0}^{1} x f(x) d x\) Given, \(a^{2} \int_{0}^{1} f(x) d x=1, \int_{0}^{1} x f(x) d x=a, \int_{0}^{1} x^{2} f(x) d=a^{2}\) \(I \stackrel{0}{=} a^{2} \times 1+a^{2}-2 a \times a\) \(I=2 a^{2}-2 a^{2}\) \(\mathrm{I}=0\)
AP EAMCET-2021-23.08.2021
Integral Calculus
86609
If \(I=\int_{0}^{1} \frac{\sin x}{\sqrt{x}} d x\) and \(J=\int_{0}^{1} \frac{\cos x}{\sqrt{x}} d x\). Then , which one of the following is true?