86738
If \(f(a+b+1-x)=f(x)\), for all \(x\), where \(a\) and \(b\) are fixed positive real numbers, then \(\frac{1}{a+b} \int_{a}^{b} x(f(x)+f(x+1)) d x\) is equal to
1 \(\int_{a+1}^{b+1} f(x+1) d x\)
2 \(\int_{a+1}^{b+1} f(x) d x\)
3 \(\int_{a-1}^{b-1} f(x+1) d x\)
4 \(\int_{a-1}^{b-1} f(x) d x\)
Explanation:
(C) : Let, the integral \(I=\frac{1}{a+b} \int_{a}^{b} x[f(x)+f(x+1)] d x \tag{i}\) On applying property \(\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\) We get, \(I=\frac{1}{a+b} \int_{a}^{b}(a+b-x)[f(a+b-x)+f(a+b-x+1)] d x\) \(I=\frac{1}{a+b} \int_{a}^{b}(a+b-x)[f(x+1)+f(x)] d x\) \(\left[\begin{array}{l}\because \mathrm{f}(\mathrm{a}+\mathrm{b}+1-x)=\mathrm{f}(\mathrm{x}) \\ \therefore \mathrm{f}(\mathrm{a}+\mathrm{b}-\mathrm{x})=\mathrm{f}(\mathrm{x}+1)\end{array}\right]\) On adding integrals equation (i) and (ii), we get - \(2 I=\frac{a+b}{a+b} \int_{a}^{b}[f(x)+f(x+1)] d x\) \(I=\frac{1}{2}\left[\int_{a}^{b} f(x) d x+\int_{a}^{b} f(x+1) d x\right]\) \(I=\frac{1}{2}\left[\int_{a}^{b} f(x) d x+\int_{a}^{b}(a+b-x+1) d x\right]\) \(I=\frac{1}{2}\left[\int_{a}^{b} f(x) d x+\int_{a}^{b} f(x) d x\right]\) \(I=\int_{a}^{b} f(x) d x\) Putting \(x=t+1\), then upper limit \(t=b-1\) and lower limit \(\mathrm{t}=\mathrm{a}-1\) and \(\mathrm{dx}=\mathrm{dt}\) So, \(\quad I=\int_{a-1}^{b-1} f(t+1) d t\) \(I=\int_{a-1}^{b-1} f(x+1) d x\)
86738
If \(f(a+b+1-x)=f(x)\), for all \(x\), where \(a\) and \(b\) are fixed positive real numbers, then \(\frac{1}{a+b} \int_{a}^{b} x(f(x)+f(x+1)) d x\) is equal to
1 \(\int_{a+1}^{b+1} f(x+1) d x\)
2 \(\int_{a+1}^{b+1} f(x) d x\)
3 \(\int_{a-1}^{b-1} f(x+1) d x\)
4 \(\int_{a-1}^{b-1} f(x) d x\)
Explanation:
(C) : Let, the integral \(I=\frac{1}{a+b} \int_{a}^{b} x[f(x)+f(x+1)] d x \tag{i}\) On applying property \(\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\) We get, \(I=\frac{1}{a+b} \int_{a}^{b}(a+b-x)[f(a+b-x)+f(a+b-x+1)] d x\) \(I=\frac{1}{a+b} \int_{a}^{b}(a+b-x)[f(x+1)+f(x)] d x\) \(\left[\begin{array}{l}\because \mathrm{f}(\mathrm{a}+\mathrm{b}+1-x)=\mathrm{f}(\mathrm{x}) \\ \therefore \mathrm{f}(\mathrm{a}+\mathrm{b}-\mathrm{x})=\mathrm{f}(\mathrm{x}+1)\end{array}\right]\) On adding integrals equation (i) and (ii), we get - \(2 I=\frac{a+b}{a+b} \int_{a}^{b}[f(x)+f(x+1)] d x\) \(I=\frac{1}{2}\left[\int_{a}^{b} f(x) d x+\int_{a}^{b} f(x+1) d x\right]\) \(I=\frac{1}{2}\left[\int_{a}^{b} f(x) d x+\int_{a}^{b}(a+b-x+1) d x\right]\) \(I=\frac{1}{2}\left[\int_{a}^{b} f(x) d x+\int_{a}^{b} f(x) d x\right]\) \(I=\int_{a}^{b} f(x) d x\) Putting \(x=t+1\), then upper limit \(t=b-1\) and lower limit \(\mathrm{t}=\mathrm{a}-1\) and \(\mathrm{dx}=\mathrm{dt}\) So, \(\quad I=\int_{a-1}^{b-1} f(t+1) d t\) \(I=\int_{a-1}^{b-1} f(x+1) d x\)
86738
If \(f(a+b+1-x)=f(x)\), for all \(x\), where \(a\) and \(b\) are fixed positive real numbers, then \(\frac{1}{a+b} \int_{a}^{b} x(f(x)+f(x+1)) d x\) is equal to
1 \(\int_{a+1}^{b+1} f(x+1) d x\)
2 \(\int_{a+1}^{b+1} f(x) d x\)
3 \(\int_{a-1}^{b-1} f(x+1) d x\)
4 \(\int_{a-1}^{b-1} f(x) d x\)
Explanation:
(C) : Let, the integral \(I=\frac{1}{a+b} \int_{a}^{b} x[f(x)+f(x+1)] d x \tag{i}\) On applying property \(\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\) We get, \(I=\frac{1}{a+b} \int_{a}^{b}(a+b-x)[f(a+b-x)+f(a+b-x+1)] d x\) \(I=\frac{1}{a+b} \int_{a}^{b}(a+b-x)[f(x+1)+f(x)] d x\) \(\left[\begin{array}{l}\because \mathrm{f}(\mathrm{a}+\mathrm{b}+1-x)=\mathrm{f}(\mathrm{x}) \\ \therefore \mathrm{f}(\mathrm{a}+\mathrm{b}-\mathrm{x})=\mathrm{f}(\mathrm{x}+1)\end{array}\right]\) On adding integrals equation (i) and (ii), we get - \(2 I=\frac{a+b}{a+b} \int_{a}^{b}[f(x)+f(x+1)] d x\) \(I=\frac{1}{2}\left[\int_{a}^{b} f(x) d x+\int_{a}^{b} f(x+1) d x\right]\) \(I=\frac{1}{2}\left[\int_{a}^{b} f(x) d x+\int_{a}^{b}(a+b-x+1) d x\right]\) \(I=\frac{1}{2}\left[\int_{a}^{b} f(x) d x+\int_{a}^{b} f(x) d x\right]\) \(I=\int_{a}^{b} f(x) d x\) Putting \(x=t+1\), then upper limit \(t=b-1\) and lower limit \(\mathrm{t}=\mathrm{a}-1\) and \(\mathrm{dx}=\mathrm{dt}\) So, \(\quad I=\int_{a-1}^{b-1} f(t+1) d t\) \(I=\int_{a-1}^{b-1} f(x+1) d x\)
86738
If \(f(a+b+1-x)=f(x)\), for all \(x\), where \(a\) and \(b\) are fixed positive real numbers, then \(\frac{1}{a+b} \int_{a}^{b} x(f(x)+f(x+1)) d x\) is equal to
1 \(\int_{a+1}^{b+1} f(x+1) d x\)
2 \(\int_{a+1}^{b+1} f(x) d x\)
3 \(\int_{a-1}^{b-1} f(x+1) d x\)
4 \(\int_{a-1}^{b-1} f(x) d x\)
Explanation:
(C) : Let, the integral \(I=\frac{1}{a+b} \int_{a}^{b} x[f(x)+f(x+1)] d x \tag{i}\) On applying property \(\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\) We get, \(I=\frac{1}{a+b} \int_{a}^{b}(a+b-x)[f(a+b-x)+f(a+b-x+1)] d x\) \(I=\frac{1}{a+b} \int_{a}^{b}(a+b-x)[f(x+1)+f(x)] d x\) \(\left[\begin{array}{l}\because \mathrm{f}(\mathrm{a}+\mathrm{b}+1-x)=\mathrm{f}(\mathrm{x}) \\ \therefore \mathrm{f}(\mathrm{a}+\mathrm{b}-\mathrm{x})=\mathrm{f}(\mathrm{x}+1)\end{array}\right]\) On adding integrals equation (i) and (ii), we get - \(2 I=\frac{a+b}{a+b} \int_{a}^{b}[f(x)+f(x+1)] d x\) \(I=\frac{1}{2}\left[\int_{a}^{b} f(x) d x+\int_{a}^{b} f(x+1) d x\right]\) \(I=\frac{1}{2}\left[\int_{a}^{b} f(x) d x+\int_{a}^{b}(a+b-x+1) d x\right]\) \(I=\frac{1}{2}\left[\int_{a}^{b} f(x) d x+\int_{a}^{b} f(x) d x\right]\) \(I=\int_{a}^{b} f(x) d x\) Putting \(x=t+1\), then upper limit \(t=b-1\) and lower limit \(\mathrm{t}=\mathrm{a}-1\) and \(\mathrm{dx}=\mathrm{dt}\) So, \(\quad I=\int_{a-1}^{b-1} f(t+1) d t\) \(I=\int_{a-1}^{b-1} f(x+1) d x\)
