Exp. (B) \(\text For (x-1) \geq 0 \Rightarrow x \geq 1\) \(\text { Also }|x-1|=1-x \text { for }(x-1)\lt 0\) \(\qquad x\lt 1, \int_0^4|x-1| d x=\int_0^1(1-x) d x+\int_1^4(x-1) d x\) \(=\left|x-\frac{x^2}{2}\right|_0^1+\left|\frac{x^2}{2}-x\right|_1^4=\left(1-\frac{1}{2}\right)+\left[\left(\frac{16}{2}-4\right)-\left(\frac{1}{2}-1\right)\right]\) \(=\frac{1}{2}+\left[4+\frac{1}{2}\right]=5\)
Karnataka CET-2011
Integral Calculus
86695
\(\int_{-2}^1[x+1] d x=\), (Where \([x]\) is greatest integer function not greater than \(x\) )
1 -1
2 0
3 1
4 2
Explanation:
Exp. (B) \(\int_{-2}^1[x+1] d x\) \(=\int_{-2}^{-1}([x]+1) d x+\int_{-1}^0([x]+1) d x+\int_0^1([x]+1) d x\) \(=\int_{-2}^{-1}(-2+1) d x+\int_{-1}^0(-1+1) d x+\int_0^1(0+1) d x\) \(=-[x]_{-2}^{-1}+0+[x]_0^1\) \(=-(-1+2)+0+(1-0)=0\)
MHT CET-2020
Integral Calculus
86696
\(\int_{-a}^{a} x^{2}\left(\frac{e^{x^{3}}-e^{-x^{3}}}{e^{x^{3}}+e^{-x^{3}}}\right) d x=\)
Exp. (B) \(\text For (x-1) \geq 0 \Rightarrow x \geq 1\) \(\text { Also }|x-1|=1-x \text { for }(x-1)\lt 0\) \(\qquad x\lt 1, \int_0^4|x-1| d x=\int_0^1(1-x) d x+\int_1^4(x-1) d x\) \(=\left|x-\frac{x^2}{2}\right|_0^1+\left|\frac{x^2}{2}-x\right|_1^4=\left(1-\frac{1}{2}\right)+\left[\left(\frac{16}{2}-4\right)-\left(\frac{1}{2}-1\right)\right]\) \(=\frac{1}{2}+\left[4+\frac{1}{2}\right]=5\)
Karnataka CET-2011
Integral Calculus
86695
\(\int_{-2}^1[x+1] d x=\), (Where \([x]\) is greatest integer function not greater than \(x\) )
1 -1
2 0
3 1
4 2
Explanation:
Exp. (B) \(\int_{-2}^1[x+1] d x\) \(=\int_{-2}^{-1}([x]+1) d x+\int_{-1}^0([x]+1) d x+\int_0^1([x]+1) d x\) \(=\int_{-2}^{-1}(-2+1) d x+\int_{-1}^0(-1+1) d x+\int_0^1(0+1) d x\) \(=-[x]_{-2}^{-1}+0+[x]_0^1\) \(=-(-1+2)+0+(1-0)=0\)
MHT CET-2020
Integral Calculus
86696
\(\int_{-a}^{a} x^{2}\left(\frac{e^{x^{3}}-e^{-x^{3}}}{e^{x^{3}}+e^{-x^{3}}}\right) d x=\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Integral Calculus
86694
The value of \(\int_0^4|x-1| d x\) is
1 \(\frac{5}{2}\)
2 5
3 4
4 1
Explanation:
Exp. (B) \(\text For (x-1) \geq 0 \Rightarrow x \geq 1\) \(\text { Also }|x-1|=1-x \text { for }(x-1)\lt 0\) \(\qquad x\lt 1, \int_0^4|x-1| d x=\int_0^1(1-x) d x+\int_1^4(x-1) d x\) \(=\left|x-\frac{x^2}{2}\right|_0^1+\left|\frac{x^2}{2}-x\right|_1^4=\left(1-\frac{1}{2}\right)+\left[\left(\frac{16}{2}-4\right)-\left(\frac{1}{2}-1\right)\right]\) \(=\frac{1}{2}+\left[4+\frac{1}{2}\right]=5\)
Karnataka CET-2011
Integral Calculus
86695
\(\int_{-2}^1[x+1] d x=\), (Where \([x]\) is greatest integer function not greater than \(x\) )
1 -1
2 0
3 1
4 2
Explanation:
Exp. (B) \(\int_{-2}^1[x+1] d x\) \(=\int_{-2}^{-1}([x]+1) d x+\int_{-1}^0([x]+1) d x+\int_0^1([x]+1) d x\) \(=\int_{-2}^{-1}(-2+1) d x+\int_{-1}^0(-1+1) d x+\int_0^1(0+1) d x\) \(=-[x]_{-2}^{-1}+0+[x]_0^1\) \(=-(-1+2)+0+(1-0)=0\)
MHT CET-2020
Integral Calculus
86696
\(\int_{-a}^{a} x^{2}\left(\frac{e^{x^{3}}-e^{-x^{3}}}{e^{x^{3}}+e^{-x^{3}}}\right) d x=\)
Exp. (B) \(\text For (x-1) \geq 0 \Rightarrow x \geq 1\) \(\text { Also }|x-1|=1-x \text { for }(x-1)\lt 0\) \(\qquad x\lt 1, \int_0^4|x-1| d x=\int_0^1(1-x) d x+\int_1^4(x-1) d x\) \(=\left|x-\frac{x^2}{2}\right|_0^1+\left|\frac{x^2}{2}-x\right|_1^4=\left(1-\frac{1}{2}\right)+\left[\left(\frac{16}{2}-4\right)-\left(\frac{1}{2}-1\right)\right]\) \(=\frac{1}{2}+\left[4+\frac{1}{2}\right]=5\)
Karnataka CET-2011
Integral Calculus
86695
\(\int_{-2}^1[x+1] d x=\), (Where \([x]\) is greatest integer function not greater than \(x\) )
1 -1
2 0
3 1
4 2
Explanation:
Exp. (B) \(\int_{-2}^1[x+1] d x\) \(=\int_{-2}^{-1}([x]+1) d x+\int_{-1}^0([x]+1) d x+\int_0^1([x]+1) d x\) \(=\int_{-2}^{-1}(-2+1) d x+\int_{-1}^0(-1+1) d x+\int_0^1(0+1) d x\) \(=-[x]_{-2}^{-1}+0+[x]_0^1\) \(=-(-1+2)+0+(1-0)=0\)
MHT CET-2020
Integral Calculus
86696
\(\int_{-a}^{a} x^{2}\left(\frac{e^{x^{3}}-e^{-x^{3}}}{e^{x^{3}}+e^{-x^{3}}}\right) d x=\)
Exp. (B) \(\text For (x-1) \geq 0 \Rightarrow x \geq 1\) \(\text { Also }|x-1|=1-x \text { for }(x-1)\lt 0\) \(\qquad x\lt 1, \int_0^4|x-1| d x=\int_0^1(1-x) d x+\int_1^4(x-1) d x\) \(=\left|x-\frac{x^2}{2}\right|_0^1+\left|\frac{x^2}{2}-x\right|_1^4=\left(1-\frac{1}{2}\right)+\left[\left(\frac{16}{2}-4\right)-\left(\frac{1}{2}-1\right)\right]\) \(=\frac{1}{2}+\left[4+\frac{1}{2}\right]=5\)
Karnataka CET-2011
Integral Calculus
86695
\(\int_{-2}^1[x+1] d x=\), (Where \([x]\) is greatest integer function not greater than \(x\) )
1 -1
2 0
3 1
4 2
Explanation:
Exp. (B) \(\int_{-2}^1[x+1] d x\) \(=\int_{-2}^{-1}([x]+1) d x+\int_{-1}^0([x]+1) d x+\int_0^1([x]+1) d x\) \(=\int_{-2}^{-1}(-2+1) d x+\int_{-1}^0(-1+1) d x+\int_0^1(0+1) d x\) \(=-[x]_{-2}^{-1}+0+[x]_0^1\) \(=-(-1+2)+0+(1-0)=0\)
MHT CET-2020
Integral Calculus
86696
\(\int_{-a}^{a} x^{2}\left(\frac{e^{x^{3}}-e^{-x^{3}}}{e^{x^{3}}+e^{-x^{3}}}\right) d x=\)