Exams Joint Entrance Examination Mathematics Definite Integration Question 80 Question 80 of 246 Medium NUMERICAL
∑ r = 1 20 ( ∣ π ( ∫ 0 r x ∣ sin π x ∣ d x ) ∣ ) \sum _{r=1}^{20}(|\sqrt{\pi (\int _{0}^{r}x|\sin \pi x|dx)}|) ∑ r = 1 20 ( ∣ π ( ∫ 0 r x ∣ sin π x ∣ d x ) ∣ )
Answer: 210
Solutions · 1 Post your solution Official solution
Let J r = ∫ 0 r x ⋅ ∣ sin π x ∣ d x {J}_{r}=\int _{0}^{r}x\cdot |\sin \pi x|dx J r = ∫ 0 r x ⋅ ∣ sin π x ∣ d x
So we need to calculate
I = ∑ r = 1 20 π J r I=\sum _{r=1}^{20}\sqrt{\pi {J}_{r}} I = ∑ r = 1 20 π J r
So,
I = π ( J 1 + J 2 + J 3 + ⋯ + J 20 ) I=\sqrt{\pi }({J}_{1}+{J}_{2}+{J}_{3}+\cdots +{J}_{20}) I = π ( J 1 + J 2 + J 3 + ⋯ + J 20 )
Let π x = t ⇒ π d x = d t \pi x=t\Rightarrow \pi dx=dt π x = t ⇒ π d x = d t
J r = ∫ 0 π r t π ∣ sin t ∣ d t π = 1 π 2 ∫ 0 π r t ∣ sin t ∣ d t {J}_{r}=\int _{0}^{\pi r}\frac{t}{\pi }|\sin t|\frac{dt}{\pi }=\frac{1}{{\pi }^{2}}\int _{0}^{\pi r}t|\sin t|dt J r = ∫ 0 π r π t ∣ sin t ∣ π d t = π 2 1 ∫ 0 π r t ∣ sin t ∣ d t
Period of ∣ sin t ∣ |\sin t| ∣ sin t ∣ is π \pi π , so break the interval
∫ 0 π r t ∣ sin t ∣ = ∫ 0 π t ∣ sin t ∣ d t + ∫ π 2 π t ∣ sin t ∣ d t + ∫ 2 π 3 π t ∣ sin t ∣ d t + ⋯ ∫ ( r − 1 ) π r π t ∣ sin t ∣ d t ∣ sin t ∣ = sin t ∣ sin t ∣ = − sin t \begin{gathered} \int _{0}^{\pi r}t|\sin t|=\int _{0}^{\pi }t|\sin t|dt+\int _{\pi }^{2\pi }t|\sin t|dt+\int _{2\pi }^{3\pi }t|\sin t|dt+\cdots \int _{(r-1)\pi }^{r\pi }t|\sin t|dt \\ \, |\sin t|=\sin t \\ \, |\sin t|=-\sin t \end{gathered} ∫ 0 π r t ∣ sin t ∣ = ∫ 0 π t ∣ sin t ∣ d t + ∫ π 2 π t ∣ sin t ∣ d t + ∫ 2 π 3 π t ∣ sin t ∣ d t + ⋯ ∫ ( r − 1 ) π r π t ∣ sin t ∣ d t ∣ sin t ∣ = sin t ∣ sin t ∣ = − sin t
∫ 0 r π t ∣ sin t ∣ d t = ∫ 0 π t sin t d t − ∫ π 2 π t sin t d t + ∫ 2 π 3 π t sin t d t − ∫ 3 π 4 π t sin t … ( − 1 ) r − 1 ∫ ( r − 1 ) π r π t sin t d t \int _{0}^{r\pi }t|\sin t|dt=\int _{0}^{\pi }t\sin tdt-\int _{\pi }^{2\pi }t\sin tdt+\int _{2\pi }^{3\pi }t\sin tdt-\int _{3\pi }^{4\pi }t\sin t\ldots (-1{)}^{r-1}\int _{(r-1)\pi }^{r\pi }t\sin tdt ∫ 0 r π t ∣ sin t ∣ d t = ∫ 0 π t sin t d t − ∫ π 2 π t sin t d t + ∫ 2 π 3 π t sin t d t − ∫ 3 π 4 π t sin t … ( − 1 ) r − 1 ∫ ( r − 1 ) π r π t sin t d t
Solving ∫ t ⋅ sin t \int t\cdot \sin t ∫ t ⋅ sin t using by parts
∫ u ⋅ v d x = u ⋅ ∫ v d x − ∫ ( u ′ ∫ v d v ) d x \int u\cdot vdx=u\cdot \int vdx-\int ({u}^{'}\int vdv)dx ∫ u ⋅ v d x = u ⋅ ∫ v d x − ∫ ( u ′ ∫ v d v ) d x
⇒ \Rightarrow ⇒ ∫ t ⋅ sin t = ( − t cos t ) − ∫ 1 ⋅ ( − cos t ) = − t cos t + sin t \int t\cdot \sin t=(-t\cos t)-\int 1\cdot (-\cos t)=-t\cos t+\sin t ∫ t ⋅ sin t = ( − t cos t ) − ∫ 1 ⋅ ( − cos t ) = − t cos t + sin t
Substituting the limits
⇒ \Rightarrow ⇒ ∫ 0 π r t ⋅ ∣ sin t ∣ d t = [ − t cos t + sin t ] π 0 − [ − t cos t + sin t ] 2 π π + [ − t cos t + sin t ] 3 π 2 π − [ − t cos t + sin t ] 4 π 3 π ⋯ ∫ ( r − 1 ) π r π t sin t ( − 1 ) r − 1 d t \begin{gathered} \int _{0}^{\pi r}t\cdot |\sin t|dt=[-t\cos t+\sin t{]}_{\pi }^{0}-[-t\cos t+\sin t{]}_{2\pi }^{\pi }+[-t\cos t+\sin t{]}_{3\pi }^{2\pi }-[-t\cos t+\sin t{]}_{4\pi }^{3\pi }\cdots \int _{(r-1)\pi }^{r\pi }t\sin t(-1{)}^{r-1}dt \end{gathered} ∫ 0 π r t ⋅ ∣ sin t ∣ d t = [ − t cos t + sin t ] π 0 − [ − t cos t + sin t ] 2 π π + [ − t cos t + sin t ] 3 π 2 π − [ − t cos t + sin t ] 4 π 3 π ⋯ ∫ ( r − 1 ) π r π t sin t ( − 1 ) r − 1 d t
= [ π + 0 − 0 ] − [ − 2 π + 0 − π + 0 ] + [ 3 π + 0 − − 2 π + 0 ] − [ − 4 π − − 3 π + 0 ] + ⋯ ∫ ( r − 1 ) π r π t sin t ( − 1 ) r − 1 d t \begin{gathered} =[\pi +0-0]-[-2\pi +0-{\pi +0}]+[3\pi +0-{-2\pi }+0]-[-4\pi -{-3\pi +0}]+\cdots \int _{(r-1)\pi }^{r\pi }t\sin t(-1{)}^{r-1}dt \end{gathered} = [ π + 0 − 0 ] − [ − 2 π + 0 − π + 0 ] + [ 3 π + 0 − − 2 π + 0 ] − [ − 4 π − − 3 π + 0 ] + ⋯ ∫ ( r − 1 ) π r π t sin t ( − 1 ) r − 1 d t
⇒ \Rightarrow ⇒ ∫ 0 r π t ∣ sin t ∣ d t = [ π + 0 ] + [ 2 π + π ] + [ 3 π + 2 π ] + [ 4 π + 3 π ] … [ r π + ( r − 1 ) π ] \int _{0}^{r\pi }t|\sin t|dt=[\pi +0]+[2\pi +\pi ]+[3\pi +2\pi ]+[4\pi +3\pi ]\ldots [r\pi +(r-1)\pi ] ∫ 0 r π t ∣ sin t ∣ d t = [ π + 0 ] + [ 2 π + π ] + [ 3 π + 2 π ] + [ 4 π + 3 π ] … [ r π + ( r − 1 ) π ]
= π + 3 π + 5 π + 7 π + … ( 2 r − 1 ) π = π ( 1 + 3 + 5 + 7 + … ( 2 r − 1 ) ) \begin{gathered} =\pi +3\pi +5\pi +7\pi +\ldots (2r-1)\pi \\ =\pi (1+3+5+7+\ldots (2r-1)) \end{gathered} = π + 3 π + 5 π + 7 π + … ( 2 r − 1 ) π = π ( 1 + 3 + 5 + 7 + … ( 2 r − 1 ))
Sum of first r r r odd numbers is r 2 {r}^{2} r 2
∫ 0 r π t ∣ sin t ∣ d t = π r 2 \int _{0}^{r\pi }t|\sin t|dt=\pi {r}^{2} ∫ 0 r π t ∣ sin t ∣ d t = π r 2
J r = 1 π 2 π r 2 = r 2 π {J}_{r}=\frac{1}{{\pi }^{2}}\pi {r}^{2}=\frac{{r}^{2}}{\pi } J r = π 2 1 π r 2 = π r 2
so,
I = ∑ r = 1 20 π ⋅ J r = ∑ r = 1 20 π ⋅ r 2 π I=\sum _{r=1}^{20}\sqrt{\pi \cdot {J}_{r}}=\sum _{r=1}^{20}\sqrt{\pi \cdot \frac{{r}^{2}}{\pi }} I = ∑ r = 1 20 π ⋅ J r = ∑ r = 1 20 π ⋅ π r 2
⇒ \Rightarrow ⇒ I = ∑ r = 1 20 r = 1 + 2 + 3 + … 20 I=\sum _{r=1}^{20}r=1+2+3+\ldots 20 I = ∑ r = 1 20 r = 1 + 2 + 3 + … 20
⇒ \Rightarrow ⇒ I = 20 × 21 2 = 210 I=\frac{20\times 21}{2}=210 I = 2 20 × 21 = 210 Answer the question to unlock