86770
The intercepts on \(x\)-axis made by tangents to the curve, \(y=\int_{0}^{x}|t| d t, x \in R\), which are parallel to the line \(y=2 x\), are equal to
1 \(\pm 2\)
2 \(\pm 3\)
3 \(\pm 4\)
4 \(\pm 1\)
Explanation:
(D) : Given, \(\frac{\mathrm{dy}}{\mathrm{dx}}=|\mathrm{x}|=2\) \(\therefore \quad \mathrm{x}= \pm 2\) We can solve for \(\mathrm{y}\), we get - \(\mathrm{y}_{1} =\int_{0}^{2}|\mathrm{t}| \mathrm{dt}=\int_{0}^{2} \mathrm{tdt}\) \(=\left[\frac{\mathrm{t}^{2}}{2}\right]_{0}^{2}=2\) And, \(\mathrm{y}_{2}=\int_{0}^{-2}|\mathrm{t}| \mathrm{dt}\) \(=-\int_{0}^{-2} \mathrm{tdt}=-\left[\frac{\mathrm{t}^{2}}{2}\right]_{0}^{-2}\) \(=-2\) Tangent are \(\mathrm{y}-2=2(\mathrm{x}-2)\) and, \(y+2=2(x+2)\) Then, the \(\mathrm{x}\) intercepts are obtained by putting \(\mathrm{y}=0\) Then, we get \(x= \pm 1\)
COMEDK-2013
Integral Calculus
86771
If \(\int_{-1}^{4} f(x) d x=4\) and \(\int_{2}^{4}(3-f(x)) d x=7\), \(\text { then } \int_{-1}^{2} f(x) d x=\)
1 -2
2 3
3 4
4 5
Explanation:
(D) : Let, \(\int_{-1}^{4} f(x) d x=4\) And, \(\int_{2}^{4}[3-f(x)] d x=7\) \(\therefore \quad 3[\mathrm{x}]_{2}^{4}-\int_{2}^{4} \mathrm{f}(\mathrm{x}) \mathrm{dx}=7\) \(\int_{2}^{4} f(x) d x=-1\) We know, \(\int_{-1}^{4} f(x) d x=\int_{-1}^{2} f(x) d x+\int_{2}^{4} f(x) d x\) \(\therefore \quad \int_{-1}^{2} f(x) d x=\int_{-1}^{4} f(x) d x-\int_{2}^{4} f(x) d x\) \(=4-(-1)=5\)
COMEDK-2019
Integral Calculus
86772
If \(\int_{0}^{b-c} f(x+c) d x=a \int_{b}^{c} f(x) d x\), then \(a=\)
1 0
2 -1
3 2
4 \(b-c\)
Explanation:
(B) : Given, \(\int_{0}^{b-c} f(x+c) d x=a \int_{b}^{c} f(x) d x\) Let, \(\mathrm{x}+\mathrm{c}=\mathrm{t}\) \(\mathrm{dx}+0=\mathrm{dt}\) \(\mathrm{dx}=\mathrm{dt}\) Also, \(x=b-c \Rightarrow t=(b-c)+c=b\) \(\mathrm{x}=0 \Rightarrow \mathrm{t}=0+\mathrm{c}\) \(\int_{0}^{b-c} f(x+c) d x=\int_{c}^{b} f(t) d t=-\int_{b}^{c} f(x) d x\) \(a \int_{b}^{c} f(x) d x=-\int_{b}^{c} f(x) d x\) \(\therefore \mathrm{a}=-1\)
COMEDK-2019
Integral Calculus
86769
The value of the integral \(\int_{0}^{\pi / 2}\left(\sin ^{100} x-\cos ^{100} x\right) d x \text { is }\)
1 \(\frac{1}{100}\)
2 \(\frac{100 !}{(100)^{100}}\)
3 \(\frac{\pi}{100}\)
4 0
Explanation:
(D) :Let, \(I=\int_{0}^{\pi / 2}\left[\sin ^{100} x-\cos ^{100} x\right] d x \tag{i}\) Apply the property, \(x=a+b-x\) \(\int_{b}^{a} x d x=\int_{b}^{a}(a+b-x) d x\) \(I=\int_{0}^{\pi / 2}\left[\sin ^{100}\left(\frac{\pi}{2}+0-x\right)-\cos ^{100}\left(\frac{\pi}{2}+0-x\right)\right] d x\) \(I=\int_{0}^{\pi / 2}\left[\cos ^{100} x-\sin ^{100} x\right] d x \tag{ii}\) On adding equation (i) and (ii), we get - \(2 I=\int_{0}^{\pi / 2}\left[\sin ^{100} x-\cos ^{100} x+\cos ^{100} x-\sin ^{100} x\right]\) \(2 I=\int_{0}^{\pi / 2} 0 d x\) \(I=0\)
Karnataka CET-2007
Integral Calculus
86764
\(\int_{0}^{1} \frac{d x}{e^{x}+e^{-x}}\) is equal to
1 \(\frac{\pi}{4}-\tan ^{-1}\) (e)
2 \(\tan ^{-1}(\mathrm{e})-\frac{\pi}{4}\)
3 \(\tan ^{-1}(\mathrm{e})+\frac{\pi}{4}\)
4 \(\tan ^{-1}(\mathrm{e})\)
Explanation:
(B) :Given, \(\int_{0}^{1} \frac{\mathrm{dx}}{\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{-\mathrm{x}}}\) Let, \(\quad I=\int_{0}^{1} \frac{d x}{e^{x}+1 / e^{x}}=\int_{0}^{1} \frac{e^{x} d x}{e^{2 x}+1}=\int_{0}^{1} \frac{e^{x} d x}{\left(e^{x}\right)^{2}+1}\) Let, \(\mathrm{e}^{\mathrm{x}}=\mathrm{t}\) \(e^{x} d x=d t\) If, \(x=0\), then \(e^{0}=t, t=1\) And, If, \(\mathrm{x}=1\) then \(\mathrm{t}=\mathrm{e}\) \(=\int_{1}^{\mathrm{e}} \frac{\mathrm{dt}}{\mathrm{t}^{2}+1}=\left[\tan ^{-1} \mathrm{t}\right]_{1}^{\mathrm{e}}=\left[\tan ^{-1}(\mathrm{e})-\tan ^{-1} 1\right]\) \(=\tan ^{-1}(\mathrm{e})-\frac{\pi}{4}\)
86770
The intercepts on \(x\)-axis made by tangents to the curve, \(y=\int_{0}^{x}|t| d t, x \in R\), which are parallel to the line \(y=2 x\), are equal to
1 \(\pm 2\)
2 \(\pm 3\)
3 \(\pm 4\)
4 \(\pm 1\)
Explanation:
(D) : Given, \(\frac{\mathrm{dy}}{\mathrm{dx}}=|\mathrm{x}|=2\) \(\therefore \quad \mathrm{x}= \pm 2\) We can solve for \(\mathrm{y}\), we get - \(\mathrm{y}_{1} =\int_{0}^{2}|\mathrm{t}| \mathrm{dt}=\int_{0}^{2} \mathrm{tdt}\) \(=\left[\frac{\mathrm{t}^{2}}{2}\right]_{0}^{2}=2\) And, \(\mathrm{y}_{2}=\int_{0}^{-2}|\mathrm{t}| \mathrm{dt}\) \(=-\int_{0}^{-2} \mathrm{tdt}=-\left[\frac{\mathrm{t}^{2}}{2}\right]_{0}^{-2}\) \(=-2\) Tangent are \(\mathrm{y}-2=2(\mathrm{x}-2)\) and, \(y+2=2(x+2)\) Then, the \(\mathrm{x}\) intercepts are obtained by putting \(\mathrm{y}=0\) Then, we get \(x= \pm 1\)
COMEDK-2013
Integral Calculus
86771
If \(\int_{-1}^{4} f(x) d x=4\) and \(\int_{2}^{4}(3-f(x)) d x=7\), \(\text { then } \int_{-1}^{2} f(x) d x=\)
1 -2
2 3
3 4
4 5
Explanation:
(D) : Let, \(\int_{-1}^{4} f(x) d x=4\) And, \(\int_{2}^{4}[3-f(x)] d x=7\) \(\therefore \quad 3[\mathrm{x}]_{2}^{4}-\int_{2}^{4} \mathrm{f}(\mathrm{x}) \mathrm{dx}=7\) \(\int_{2}^{4} f(x) d x=-1\) We know, \(\int_{-1}^{4} f(x) d x=\int_{-1}^{2} f(x) d x+\int_{2}^{4} f(x) d x\) \(\therefore \quad \int_{-1}^{2} f(x) d x=\int_{-1}^{4} f(x) d x-\int_{2}^{4} f(x) d x\) \(=4-(-1)=5\)
COMEDK-2019
Integral Calculus
86772
If \(\int_{0}^{b-c} f(x+c) d x=a \int_{b}^{c} f(x) d x\), then \(a=\)
1 0
2 -1
3 2
4 \(b-c\)
Explanation:
(B) : Given, \(\int_{0}^{b-c} f(x+c) d x=a \int_{b}^{c} f(x) d x\) Let, \(\mathrm{x}+\mathrm{c}=\mathrm{t}\) \(\mathrm{dx}+0=\mathrm{dt}\) \(\mathrm{dx}=\mathrm{dt}\) Also, \(x=b-c \Rightarrow t=(b-c)+c=b\) \(\mathrm{x}=0 \Rightarrow \mathrm{t}=0+\mathrm{c}\) \(\int_{0}^{b-c} f(x+c) d x=\int_{c}^{b} f(t) d t=-\int_{b}^{c} f(x) d x\) \(a \int_{b}^{c} f(x) d x=-\int_{b}^{c} f(x) d x\) \(\therefore \mathrm{a}=-1\)
COMEDK-2019
Integral Calculus
86769
The value of the integral \(\int_{0}^{\pi / 2}\left(\sin ^{100} x-\cos ^{100} x\right) d x \text { is }\)
1 \(\frac{1}{100}\)
2 \(\frac{100 !}{(100)^{100}}\)
3 \(\frac{\pi}{100}\)
4 0
Explanation:
(D) :Let, \(I=\int_{0}^{\pi / 2}\left[\sin ^{100} x-\cos ^{100} x\right] d x \tag{i}\) Apply the property, \(x=a+b-x\) \(\int_{b}^{a} x d x=\int_{b}^{a}(a+b-x) d x\) \(I=\int_{0}^{\pi / 2}\left[\sin ^{100}\left(\frac{\pi}{2}+0-x\right)-\cos ^{100}\left(\frac{\pi}{2}+0-x\right)\right] d x\) \(I=\int_{0}^{\pi / 2}\left[\cos ^{100} x-\sin ^{100} x\right] d x \tag{ii}\) On adding equation (i) and (ii), we get - \(2 I=\int_{0}^{\pi / 2}\left[\sin ^{100} x-\cos ^{100} x+\cos ^{100} x-\sin ^{100} x\right]\) \(2 I=\int_{0}^{\pi / 2} 0 d x\) \(I=0\)
Karnataka CET-2007
Integral Calculus
86764
\(\int_{0}^{1} \frac{d x}{e^{x}+e^{-x}}\) is equal to
1 \(\frac{\pi}{4}-\tan ^{-1}\) (e)
2 \(\tan ^{-1}(\mathrm{e})-\frac{\pi}{4}\)
3 \(\tan ^{-1}(\mathrm{e})+\frac{\pi}{4}\)
4 \(\tan ^{-1}(\mathrm{e})\)
Explanation:
(B) :Given, \(\int_{0}^{1} \frac{\mathrm{dx}}{\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{-\mathrm{x}}}\) Let, \(\quad I=\int_{0}^{1} \frac{d x}{e^{x}+1 / e^{x}}=\int_{0}^{1} \frac{e^{x} d x}{e^{2 x}+1}=\int_{0}^{1} \frac{e^{x} d x}{\left(e^{x}\right)^{2}+1}\) Let, \(\mathrm{e}^{\mathrm{x}}=\mathrm{t}\) \(e^{x} d x=d t\) If, \(x=0\), then \(e^{0}=t, t=1\) And, If, \(\mathrm{x}=1\) then \(\mathrm{t}=\mathrm{e}\) \(=\int_{1}^{\mathrm{e}} \frac{\mathrm{dt}}{\mathrm{t}^{2}+1}=\left[\tan ^{-1} \mathrm{t}\right]_{1}^{\mathrm{e}}=\left[\tan ^{-1}(\mathrm{e})-\tan ^{-1} 1\right]\) \(=\tan ^{-1}(\mathrm{e})-\frac{\pi}{4}\)
86770
The intercepts on \(x\)-axis made by tangents to the curve, \(y=\int_{0}^{x}|t| d t, x \in R\), which are parallel to the line \(y=2 x\), are equal to
1 \(\pm 2\)
2 \(\pm 3\)
3 \(\pm 4\)
4 \(\pm 1\)
Explanation:
(D) : Given, \(\frac{\mathrm{dy}}{\mathrm{dx}}=|\mathrm{x}|=2\) \(\therefore \quad \mathrm{x}= \pm 2\) We can solve for \(\mathrm{y}\), we get - \(\mathrm{y}_{1} =\int_{0}^{2}|\mathrm{t}| \mathrm{dt}=\int_{0}^{2} \mathrm{tdt}\) \(=\left[\frac{\mathrm{t}^{2}}{2}\right]_{0}^{2}=2\) And, \(\mathrm{y}_{2}=\int_{0}^{-2}|\mathrm{t}| \mathrm{dt}\) \(=-\int_{0}^{-2} \mathrm{tdt}=-\left[\frac{\mathrm{t}^{2}}{2}\right]_{0}^{-2}\) \(=-2\) Tangent are \(\mathrm{y}-2=2(\mathrm{x}-2)\) and, \(y+2=2(x+2)\) Then, the \(\mathrm{x}\) intercepts are obtained by putting \(\mathrm{y}=0\) Then, we get \(x= \pm 1\)
COMEDK-2013
Integral Calculus
86771
If \(\int_{-1}^{4} f(x) d x=4\) and \(\int_{2}^{4}(3-f(x)) d x=7\), \(\text { then } \int_{-1}^{2} f(x) d x=\)
1 -2
2 3
3 4
4 5
Explanation:
(D) : Let, \(\int_{-1}^{4} f(x) d x=4\) And, \(\int_{2}^{4}[3-f(x)] d x=7\) \(\therefore \quad 3[\mathrm{x}]_{2}^{4}-\int_{2}^{4} \mathrm{f}(\mathrm{x}) \mathrm{dx}=7\) \(\int_{2}^{4} f(x) d x=-1\) We know, \(\int_{-1}^{4} f(x) d x=\int_{-1}^{2} f(x) d x+\int_{2}^{4} f(x) d x\) \(\therefore \quad \int_{-1}^{2} f(x) d x=\int_{-1}^{4} f(x) d x-\int_{2}^{4} f(x) d x\) \(=4-(-1)=5\)
COMEDK-2019
Integral Calculus
86772
If \(\int_{0}^{b-c} f(x+c) d x=a \int_{b}^{c} f(x) d x\), then \(a=\)
1 0
2 -1
3 2
4 \(b-c\)
Explanation:
(B) : Given, \(\int_{0}^{b-c} f(x+c) d x=a \int_{b}^{c} f(x) d x\) Let, \(\mathrm{x}+\mathrm{c}=\mathrm{t}\) \(\mathrm{dx}+0=\mathrm{dt}\) \(\mathrm{dx}=\mathrm{dt}\) Also, \(x=b-c \Rightarrow t=(b-c)+c=b\) \(\mathrm{x}=0 \Rightarrow \mathrm{t}=0+\mathrm{c}\) \(\int_{0}^{b-c} f(x+c) d x=\int_{c}^{b} f(t) d t=-\int_{b}^{c} f(x) d x\) \(a \int_{b}^{c} f(x) d x=-\int_{b}^{c} f(x) d x\) \(\therefore \mathrm{a}=-1\)
COMEDK-2019
Integral Calculus
86769
The value of the integral \(\int_{0}^{\pi / 2}\left(\sin ^{100} x-\cos ^{100} x\right) d x \text { is }\)
1 \(\frac{1}{100}\)
2 \(\frac{100 !}{(100)^{100}}\)
3 \(\frac{\pi}{100}\)
4 0
Explanation:
(D) :Let, \(I=\int_{0}^{\pi / 2}\left[\sin ^{100} x-\cos ^{100} x\right] d x \tag{i}\) Apply the property, \(x=a+b-x\) \(\int_{b}^{a} x d x=\int_{b}^{a}(a+b-x) d x\) \(I=\int_{0}^{\pi / 2}\left[\sin ^{100}\left(\frac{\pi}{2}+0-x\right)-\cos ^{100}\left(\frac{\pi}{2}+0-x\right)\right] d x\) \(I=\int_{0}^{\pi / 2}\left[\cos ^{100} x-\sin ^{100} x\right] d x \tag{ii}\) On adding equation (i) and (ii), we get - \(2 I=\int_{0}^{\pi / 2}\left[\sin ^{100} x-\cos ^{100} x+\cos ^{100} x-\sin ^{100} x\right]\) \(2 I=\int_{0}^{\pi / 2} 0 d x\) \(I=0\)
Karnataka CET-2007
Integral Calculus
86764
\(\int_{0}^{1} \frac{d x}{e^{x}+e^{-x}}\) is equal to
1 \(\frac{\pi}{4}-\tan ^{-1}\) (e)
2 \(\tan ^{-1}(\mathrm{e})-\frac{\pi}{4}\)
3 \(\tan ^{-1}(\mathrm{e})+\frac{\pi}{4}\)
4 \(\tan ^{-1}(\mathrm{e})\)
Explanation:
(B) :Given, \(\int_{0}^{1} \frac{\mathrm{dx}}{\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{-\mathrm{x}}}\) Let, \(\quad I=\int_{0}^{1} \frac{d x}{e^{x}+1 / e^{x}}=\int_{0}^{1} \frac{e^{x} d x}{e^{2 x}+1}=\int_{0}^{1} \frac{e^{x} d x}{\left(e^{x}\right)^{2}+1}\) Let, \(\mathrm{e}^{\mathrm{x}}=\mathrm{t}\) \(e^{x} d x=d t\) If, \(x=0\), then \(e^{0}=t, t=1\) And, If, \(\mathrm{x}=1\) then \(\mathrm{t}=\mathrm{e}\) \(=\int_{1}^{\mathrm{e}} \frac{\mathrm{dt}}{\mathrm{t}^{2}+1}=\left[\tan ^{-1} \mathrm{t}\right]_{1}^{\mathrm{e}}=\left[\tan ^{-1}(\mathrm{e})-\tan ^{-1} 1\right]\) \(=\tan ^{-1}(\mathrm{e})-\frac{\pi}{4}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Integral Calculus
86770
The intercepts on \(x\)-axis made by tangents to the curve, \(y=\int_{0}^{x}|t| d t, x \in R\), which are parallel to the line \(y=2 x\), are equal to
1 \(\pm 2\)
2 \(\pm 3\)
3 \(\pm 4\)
4 \(\pm 1\)
Explanation:
(D) : Given, \(\frac{\mathrm{dy}}{\mathrm{dx}}=|\mathrm{x}|=2\) \(\therefore \quad \mathrm{x}= \pm 2\) We can solve for \(\mathrm{y}\), we get - \(\mathrm{y}_{1} =\int_{0}^{2}|\mathrm{t}| \mathrm{dt}=\int_{0}^{2} \mathrm{tdt}\) \(=\left[\frac{\mathrm{t}^{2}}{2}\right]_{0}^{2}=2\) And, \(\mathrm{y}_{2}=\int_{0}^{-2}|\mathrm{t}| \mathrm{dt}\) \(=-\int_{0}^{-2} \mathrm{tdt}=-\left[\frac{\mathrm{t}^{2}}{2}\right]_{0}^{-2}\) \(=-2\) Tangent are \(\mathrm{y}-2=2(\mathrm{x}-2)\) and, \(y+2=2(x+2)\) Then, the \(\mathrm{x}\) intercepts are obtained by putting \(\mathrm{y}=0\) Then, we get \(x= \pm 1\)
COMEDK-2013
Integral Calculus
86771
If \(\int_{-1}^{4} f(x) d x=4\) and \(\int_{2}^{4}(3-f(x)) d x=7\), \(\text { then } \int_{-1}^{2} f(x) d x=\)
1 -2
2 3
3 4
4 5
Explanation:
(D) : Let, \(\int_{-1}^{4} f(x) d x=4\) And, \(\int_{2}^{4}[3-f(x)] d x=7\) \(\therefore \quad 3[\mathrm{x}]_{2}^{4}-\int_{2}^{4} \mathrm{f}(\mathrm{x}) \mathrm{dx}=7\) \(\int_{2}^{4} f(x) d x=-1\) We know, \(\int_{-1}^{4} f(x) d x=\int_{-1}^{2} f(x) d x+\int_{2}^{4} f(x) d x\) \(\therefore \quad \int_{-1}^{2} f(x) d x=\int_{-1}^{4} f(x) d x-\int_{2}^{4} f(x) d x\) \(=4-(-1)=5\)
COMEDK-2019
Integral Calculus
86772
If \(\int_{0}^{b-c} f(x+c) d x=a \int_{b}^{c} f(x) d x\), then \(a=\)
1 0
2 -1
3 2
4 \(b-c\)
Explanation:
(B) : Given, \(\int_{0}^{b-c} f(x+c) d x=a \int_{b}^{c} f(x) d x\) Let, \(\mathrm{x}+\mathrm{c}=\mathrm{t}\) \(\mathrm{dx}+0=\mathrm{dt}\) \(\mathrm{dx}=\mathrm{dt}\) Also, \(x=b-c \Rightarrow t=(b-c)+c=b\) \(\mathrm{x}=0 \Rightarrow \mathrm{t}=0+\mathrm{c}\) \(\int_{0}^{b-c} f(x+c) d x=\int_{c}^{b} f(t) d t=-\int_{b}^{c} f(x) d x\) \(a \int_{b}^{c} f(x) d x=-\int_{b}^{c} f(x) d x\) \(\therefore \mathrm{a}=-1\)
COMEDK-2019
Integral Calculus
86769
The value of the integral \(\int_{0}^{\pi / 2}\left(\sin ^{100} x-\cos ^{100} x\right) d x \text { is }\)
1 \(\frac{1}{100}\)
2 \(\frac{100 !}{(100)^{100}}\)
3 \(\frac{\pi}{100}\)
4 0
Explanation:
(D) :Let, \(I=\int_{0}^{\pi / 2}\left[\sin ^{100} x-\cos ^{100} x\right] d x \tag{i}\) Apply the property, \(x=a+b-x\) \(\int_{b}^{a} x d x=\int_{b}^{a}(a+b-x) d x\) \(I=\int_{0}^{\pi / 2}\left[\sin ^{100}\left(\frac{\pi}{2}+0-x\right)-\cos ^{100}\left(\frac{\pi}{2}+0-x\right)\right] d x\) \(I=\int_{0}^{\pi / 2}\left[\cos ^{100} x-\sin ^{100} x\right] d x \tag{ii}\) On adding equation (i) and (ii), we get - \(2 I=\int_{0}^{\pi / 2}\left[\sin ^{100} x-\cos ^{100} x+\cos ^{100} x-\sin ^{100} x\right]\) \(2 I=\int_{0}^{\pi / 2} 0 d x\) \(I=0\)
Karnataka CET-2007
Integral Calculus
86764
\(\int_{0}^{1} \frac{d x}{e^{x}+e^{-x}}\) is equal to
1 \(\frac{\pi}{4}-\tan ^{-1}\) (e)
2 \(\tan ^{-1}(\mathrm{e})-\frac{\pi}{4}\)
3 \(\tan ^{-1}(\mathrm{e})+\frac{\pi}{4}\)
4 \(\tan ^{-1}(\mathrm{e})\)
Explanation:
(B) :Given, \(\int_{0}^{1} \frac{\mathrm{dx}}{\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{-\mathrm{x}}}\) Let, \(\quad I=\int_{0}^{1} \frac{d x}{e^{x}+1 / e^{x}}=\int_{0}^{1} \frac{e^{x} d x}{e^{2 x}+1}=\int_{0}^{1} \frac{e^{x} d x}{\left(e^{x}\right)^{2}+1}\) Let, \(\mathrm{e}^{\mathrm{x}}=\mathrm{t}\) \(e^{x} d x=d t\) If, \(x=0\), then \(e^{0}=t, t=1\) And, If, \(\mathrm{x}=1\) then \(\mathrm{t}=\mathrm{e}\) \(=\int_{1}^{\mathrm{e}} \frac{\mathrm{dt}}{\mathrm{t}^{2}+1}=\left[\tan ^{-1} \mathrm{t}\right]_{1}^{\mathrm{e}}=\left[\tan ^{-1}(\mathrm{e})-\tan ^{-1} 1\right]\) \(=\tan ^{-1}(\mathrm{e})-\frac{\pi}{4}\)
86770
The intercepts on \(x\)-axis made by tangents to the curve, \(y=\int_{0}^{x}|t| d t, x \in R\), which are parallel to the line \(y=2 x\), are equal to
1 \(\pm 2\)
2 \(\pm 3\)
3 \(\pm 4\)
4 \(\pm 1\)
Explanation:
(D) : Given, \(\frac{\mathrm{dy}}{\mathrm{dx}}=|\mathrm{x}|=2\) \(\therefore \quad \mathrm{x}= \pm 2\) We can solve for \(\mathrm{y}\), we get - \(\mathrm{y}_{1} =\int_{0}^{2}|\mathrm{t}| \mathrm{dt}=\int_{0}^{2} \mathrm{tdt}\) \(=\left[\frac{\mathrm{t}^{2}}{2}\right]_{0}^{2}=2\) And, \(\mathrm{y}_{2}=\int_{0}^{-2}|\mathrm{t}| \mathrm{dt}\) \(=-\int_{0}^{-2} \mathrm{tdt}=-\left[\frac{\mathrm{t}^{2}}{2}\right]_{0}^{-2}\) \(=-2\) Tangent are \(\mathrm{y}-2=2(\mathrm{x}-2)\) and, \(y+2=2(x+2)\) Then, the \(\mathrm{x}\) intercepts are obtained by putting \(\mathrm{y}=0\) Then, we get \(x= \pm 1\)
COMEDK-2013
Integral Calculus
86771
If \(\int_{-1}^{4} f(x) d x=4\) and \(\int_{2}^{4}(3-f(x)) d x=7\), \(\text { then } \int_{-1}^{2} f(x) d x=\)
1 -2
2 3
3 4
4 5
Explanation:
(D) : Let, \(\int_{-1}^{4} f(x) d x=4\) And, \(\int_{2}^{4}[3-f(x)] d x=7\) \(\therefore \quad 3[\mathrm{x}]_{2}^{4}-\int_{2}^{4} \mathrm{f}(\mathrm{x}) \mathrm{dx}=7\) \(\int_{2}^{4} f(x) d x=-1\) We know, \(\int_{-1}^{4} f(x) d x=\int_{-1}^{2} f(x) d x+\int_{2}^{4} f(x) d x\) \(\therefore \quad \int_{-1}^{2} f(x) d x=\int_{-1}^{4} f(x) d x-\int_{2}^{4} f(x) d x\) \(=4-(-1)=5\)
COMEDK-2019
Integral Calculus
86772
If \(\int_{0}^{b-c} f(x+c) d x=a \int_{b}^{c} f(x) d x\), then \(a=\)
1 0
2 -1
3 2
4 \(b-c\)
Explanation:
(B) : Given, \(\int_{0}^{b-c} f(x+c) d x=a \int_{b}^{c} f(x) d x\) Let, \(\mathrm{x}+\mathrm{c}=\mathrm{t}\) \(\mathrm{dx}+0=\mathrm{dt}\) \(\mathrm{dx}=\mathrm{dt}\) Also, \(x=b-c \Rightarrow t=(b-c)+c=b\) \(\mathrm{x}=0 \Rightarrow \mathrm{t}=0+\mathrm{c}\) \(\int_{0}^{b-c} f(x+c) d x=\int_{c}^{b} f(t) d t=-\int_{b}^{c} f(x) d x\) \(a \int_{b}^{c} f(x) d x=-\int_{b}^{c} f(x) d x\) \(\therefore \mathrm{a}=-1\)
COMEDK-2019
Integral Calculus
86769
The value of the integral \(\int_{0}^{\pi / 2}\left(\sin ^{100} x-\cos ^{100} x\right) d x \text { is }\)
1 \(\frac{1}{100}\)
2 \(\frac{100 !}{(100)^{100}}\)
3 \(\frac{\pi}{100}\)
4 0
Explanation:
(D) :Let, \(I=\int_{0}^{\pi / 2}\left[\sin ^{100} x-\cos ^{100} x\right] d x \tag{i}\) Apply the property, \(x=a+b-x\) \(\int_{b}^{a} x d x=\int_{b}^{a}(a+b-x) d x\) \(I=\int_{0}^{\pi / 2}\left[\sin ^{100}\left(\frac{\pi}{2}+0-x\right)-\cos ^{100}\left(\frac{\pi}{2}+0-x\right)\right] d x\) \(I=\int_{0}^{\pi / 2}\left[\cos ^{100} x-\sin ^{100} x\right] d x \tag{ii}\) On adding equation (i) and (ii), we get - \(2 I=\int_{0}^{\pi / 2}\left[\sin ^{100} x-\cos ^{100} x+\cos ^{100} x-\sin ^{100} x\right]\) \(2 I=\int_{0}^{\pi / 2} 0 d x\) \(I=0\)
Karnataka CET-2007
Integral Calculus
86764
\(\int_{0}^{1} \frac{d x}{e^{x}+e^{-x}}\) is equal to
1 \(\frac{\pi}{4}-\tan ^{-1}\) (e)
2 \(\tan ^{-1}(\mathrm{e})-\frac{\pi}{4}\)
3 \(\tan ^{-1}(\mathrm{e})+\frac{\pi}{4}\)
4 \(\tan ^{-1}(\mathrm{e})\)
Explanation:
(B) :Given, \(\int_{0}^{1} \frac{\mathrm{dx}}{\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{-\mathrm{x}}}\) Let, \(\quad I=\int_{0}^{1} \frac{d x}{e^{x}+1 / e^{x}}=\int_{0}^{1} \frac{e^{x} d x}{e^{2 x}+1}=\int_{0}^{1} \frac{e^{x} d x}{\left(e^{x}\right)^{2}+1}\) Let, \(\mathrm{e}^{\mathrm{x}}=\mathrm{t}\) \(e^{x} d x=d t\) If, \(x=0\), then \(e^{0}=t, t=1\) And, If, \(\mathrm{x}=1\) then \(\mathrm{t}=\mathrm{e}\) \(=\int_{1}^{\mathrm{e}} \frac{\mathrm{dt}}{\mathrm{t}^{2}+1}=\left[\tan ^{-1} \mathrm{t}\right]_{1}^{\mathrm{e}}=\left[\tan ^{-1}(\mathrm{e})-\tan ^{-1} 1\right]\) \(=\tan ^{-1}(\mathrm{e})-\frac{\pi}{4}\)