86559
\(\int_{-1}^{2}\left|x^{3}-x\right| d x\) is equal to
1 11
2 4
3 \(\frac{11}{4}\)
4 \(\frac{4}{11}\)
Explanation:
(C) : We have, \(\int_{-1}^{2}\left|x^{3}-x\right| d x\) \(=\int_{-1}^{0}\left(x^{3}-x\right) d x-\int_{0}^{1}\left(x^{3}-x\right) d x+\int_{1}^{2}\left(x^{3}-x\right) d x\) \(=\left[\frac{x^{4}}{4}-\frac{x^{2}}{2}\right]_{-1}^{0}-\left[\frac{x^{4}}{4}-\frac{x^{2}}{2}\right]_{0}^{1}+\left[\frac{x^{4}}{4}-\frac{x^{2}}{2}\right]_{1}^{2}\) \(=\left[(0-0)-\left(\frac{1}{4}-\frac{1}{2}\right)\right]-\left[\left(\frac{1}{4}-\frac{1}{2}\right)-(0-0)\right]+\left[(4-2)-\left(\frac{1}{4}-\frac{1}{2}\right)\right]\) \(=\frac{1}{4}+\frac{1}{4}+2+\frac{1}{4}=\frac{11}{4}\)
CG PET-2013
Integral Calculus
86567
The value of \(\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{(2+3 \sin x)}{\sin x(1+\cos x)} d x\) is equal to
(C): Given, \(I=\int_{2}^{4}\{|x-2|+|x-3|\} d x\) \(|x-2|=\left\{\begin{array}{l}x-2, x \geq 2 \\ -(x-2), x\lt 2\end{array}\right.\), \(|x-3|=\left\{\begin{array}{l}x-3, \quad x \geq 3 \\ -(x-3), \quad x\lt 3\end{array}\right.\) \(=\int_{1}^{3}(|x-2|+|x-3|) d x+\int_{3}^{4}(|x-2|+|x-3|) d x\) \(=\int_{2}^{3}((x-2)+(-(x-3))) d x+\int_{3}^{4}((x-2)+(x-3)) d x\) \(=\int_{2}^{3}[x-2-x+3] d x+\int_{3}^{4}(x-2+x-3) d x\) \(=\int_{2}^{3} 1 d x+\int_{3}^{4}(2 x-5) d x\) \(=[x]_{2}^{3}+\left[\frac{2 x^{2}}{2}-5 x\right]_{3}^{4}\) \(=(3-2)+\left(4^{2}-5(4)\right)-\left(3^{2}-5(3)\right)\) \(=1+(16-20)-(9-15)\) \(=1+(-4)-(-6)=1-4+6=7-4=3\)
AP EAMCET-2021-19.08.2021
Integral Calculus
86576
The value of the integral \(\int_{-2}^{0} \frac{d x}{\sqrt{12-x^{2}-4 x}}\) is
86559
\(\int_{-1}^{2}\left|x^{3}-x\right| d x\) is equal to
1 11
2 4
3 \(\frac{11}{4}\)
4 \(\frac{4}{11}\)
Explanation:
(C) : We have, \(\int_{-1}^{2}\left|x^{3}-x\right| d x\) \(=\int_{-1}^{0}\left(x^{3}-x\right) d x-\int_{0}^{1}\left(x^{3}-x\right) d x+\int_{1}^{2}\left(x^{3}-x\right) d x\) \(=\left[\frac{x^{4}}{4}-\frac{x^{2}}{2}\right]_{-1}^{0}-\left[\frac{x^{4}}{4}-\frac{x^{2}}{2}\right]_{0}^{1}+\left[\frac{x^{4}}{4}-\frac{x^{2}}{2}\right]_{1}^{2}\) \(=\left[(0-0)-\left(\frac{1}{4}-\frac{1}{2}\right)\right]-\left[\left(\frac{1}{4}-\frac{1}{2}\right)-(0-0)\right]+\left[(4-2)-\left(\frac{1}{4}-\frac{1}{2}\right)\right]\) \(=\frac{1}{4}+\frac{1}{4}+2+\frac{1}{4}=\frac{11}{4}\)
CG PET-2013
Integral Calculus
86567
The value of \(\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{(2+3 \sin x)}{\sin x(1+\cos x)} d x\) is equal to
(C): Given, \(I=\int_{2}^{4}\{|x-2|+|x-3|\} d x\) \(|x-2|=\left\{\begin{array}{l}x-2, x \geq 2 \\ -(x-2), x\lt 2\end{array}\right.\), \(|x-3|=\left\{\begin{array}{l}x-3, \quad x \geq 3 \\ -(x-3), \quad x\lt 3\end{array}\right.\) \(=\int_{1}^{3}(|x-2|+|x-3|) d x+\int_{3}^{4}(|x-2|+|x-3|) d x\) \(=\int_{2}^{3}((x-2)+(-(x-3))) d x+\int_{3}^{4}((x-2)+(x-3)) d x\) \(=\int_{2}^{3}[x-2-x+3] d x+\int_{3}^{4}(x-2+x-3) d x\) \(=\int_{2}^{3} 1 d x+\int_{3}^{4}(2 x-5) d x\) \(=[x]_{2}^{3}+\left[\frac{2 x^{2}}{2}-5 x\right]_{3}^{4}\) \(=(3-2)+\left(4^{2}-5(4)\right)-\left(3^{2}-5(3)\right)\) \(=1+(16-20)-(9-15)\) \(=1+(-4)-(-6)=1-4+6=7-4=3\)
AP EAMCET-2021-19.08.2021
Integral Calculus
86576
The value of the integral \(\int_{-2}^{0} \frac{d x}{\sqrt{12-x^{2}-4 x}}\) is
86559
\(\int_{-1}^{2}\left|x^{3}-x\right| d x\) is equal to
1 11
2 4
3 \(\frac{11}{4}\)
4 \(\frac{4}{11}\)
Explanation:
(C) : We have, \(\int_{-1}^{2}\left|x^{3}-x\right| d x\) \(=\int_{-1}^{0}\left(x^{3}-x\right) d x-\int_{0}^{1}\left(x^{3}-x\right) d x+\int_{1}^{2}\left(x^{3}-x\right) d x\) \(=\left[\frac{x^{4}}{4}-\frac{x^{2}}{2}\right]_{-1}^{0}-\left[\frac{x^{4}}{4}-\frac{x^{2}}{2}\right]_{0}^{1}+\left[\frac{x^{4}}{4}-\frac{x^{2}}{2}\right]_{1}^{2}\) \(=\left[(0-0)-\left(\frac{1}{4}-\frac{1}{2}\right)\right]-\left[\left(\frac{1}{4}-\frac{1}{2}\right)-(0-0)\right]+\left[(4-2)-\left(\frac{1}{4}-\frac{1}{2}\right)\right]\) \(=\frac{1}{4}+\frac{1}{4}+2+\frac{1}{4}=\frac{11}{4}\)
CG PET-2013
Integral Calculus
86567
The value of \(\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{(2+3 \sin x)}{\sin x(1+\cos x)} d x\) is equal to
(C): Given, \(I=\int_{2}^{4}\{|x-2|+|x-3|\} d x\) \(|x-2|=\left\{\begin{array}{l}x-2, x \geq 2 \\ -(x-2), x\lt 2\end{array}\right.\), \(|x-3|=\left\{\begin{array}{l}x-3, \quad x \geq 3 \\ -(x-3), \quad x\lt 3\end{array}\right.\) \(=\int_{1}^{3}(|x-2|+|x-3|) d x+\int_{3}^{4}(|x-2|+|x-3|) d x\) \(=\int_{2}^{3}((x-2)+(-(x-3))) d x+\int_{3}^{4}((x-2)+(x-3)) d x\) \(=\int_{2}^{3}[x-2-x+3] d x+\int_{3}^{4}(x-2+x-3) d x\) \(=\int_{2}^{3} 1 d x+\int_{3}^{4}(2 x-5) d x\) \(=[x]_{2}^{3}+\left[\frac{2 x^{2}}{2}-5 x\right]_{3}^{4}\) \(=(3-2)+\left(4^{2}-5(4)\right)-\left(3^{2}-5(3)\right)\) \(=1+(16-20)-(9-15)\) \(=1+(-4)-(-6)=1-4+6=7-4=3\)
AP EAMCET-2021-19.08.2021
Integral Calculus
86576
The value of the integral \(\int_{-2}^{0} \frac{d x}{\sqrt{12-x^{2}-4 x}}\) is
86559
\(\int_{-1}^{2}\left|x^{3}-x\right| d x\) is equal to
1 11
2 4
3 \(\frac{11}{4}\)
4 \(\frac{4}{11}\)
Explanation:
(C) : We have, \(\int_{-1}^{2}\left|x^{3}-x\right| d x\) \(=\int_{-1}^{0}\left(x^{3}-x\right) d x-\int_{0}^{1}\left(x^{3}-x\right) d x+\int_{1}^{2}\left(x^{3}-x\right) d x\) \(=\left[\frac{x^{4}}{4}-\frac{x^{2}}{2}\right]_{-1}^{0}-\left[\frac{x^{4}}{4}-\frac{x^{2}}{2}\right]_{0}^{1}+\left[\frac{x^{4}}{4}-\frac{x^{2}}{2}\right]_{1}^{2}\) \(=\left[(0-0)-\left(\frac{1}{4}-\frac{1}{2}\right)\right]-\left[\left(\frac{1}{4}-\frac{1}{2}\right)-(0-0)\right]+\left[(4-2)-\left(\frac{1}{4}-\frac{1}{2}\right)\right]\) \(=\frac{1}{4}+\frac{1}{4}+2+\frac{1}{4}=\frac{11}{4}\)
CG PET-2013
Integral Calculus
86567
The value of \(\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{(2+3 \sin x)}{\sin x(1+\cos x)} d x\) is equal to
(C): Given, \(I=\int_{2}^{4}\{|x-2|+|x-3|\} d x\) \(|x-2|=\left\{\begin{array}{l}x-2, x \geq 2 \\ -(x-2), x\lt 2\end{array}\right.\), \(|x-3|=\left\{\begin{array}{l}x-3, \quad x \geq 3 \\ -(x-3), \quad x\lt 3\end{array}\right.\) \(=\int_{1}^{3}(|x-2|+|x-3|) d x+\int_{3}^{4}(|x-2|+|x-3|) d x\) \(=\int_{2}^{3}((x-2)+(-(x-3))) d x+\int_{3}^{4}((x-2)+(x-3)) d x\) \(=\int_{2}^{3}[x-2-x+3] d x+\int_{3}^{4}(x-2+x-3) d x\) \(=\int_{2}^{3} 1 d x+\int_{3}^{4}(2 x-5) d x\) \(=[x]_{2}^{3}+\left[\frac{2 x^{2}}{2}-5 x\right]_{3}^{4}\) \(=(3-2)+\left(4^{2}-5(4)\right)-\left(3^{2}-5(3)\right)\) \(=1+(16-20)-(9-15)\) \(=1+(-4)-(-6)=1-4+6=7-4=3\)
AP EAMCET-2021-19.08.2021
Integral Calculus
86576
The value of the integral \(\int_{-2}^{0} \frac{d x}{\sqrt{12-x^{2}-4 x}}\) is
