Class 12 Maths NCERT Solutions for Chapter 8 Application of Integrals Miscellaneous Exercise
![Class 12 Maths NCERT Solutions for Chapter 8 Application of Integrals Miscellaneous Exercise Class 12 Maths NCERT Solutions for Chapter 8 Application of Integrals Miscellaneous Exercise](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgEJm5xpPra_Jyuvtkgy3Mtjrs_UExyWBWc3VX6iIiGk9SndPqezY305yxy5C8UeD_RoF0KhAUXP6fIHFoS8eZsoZsyDA3yXq7jUO5dCoTmtov9DBh9ZQB6osIf-mpw9ygsgbEWJFxX7SFH72x5jCQJZISJj9jMtFAeNAlI1qZNs0dQMlujFADgI9p8/w647-h306-rw/ncert-solutions-for-class-12-maths-application-of-integrals-miscellaneous-exercise.jpg)
Integrals Miscellaneous Exercise Solutions
1. Find the area under the given curves and given lines:
(i) y = x2, x = 1, x = 2 and x-axis
(ii) y = x4, x = 1, x = 5 and x –axis
Solution
(i) The required area is represented by the shaded area ADCBA as
Area ADCBA = ∫12 y dx
= ∫12 x2 dx
= 8/3 - 1/3
= 7/3 units
(ii) The required area is represented by the shaded area ADCBA as
Area ADCBA = ∫15 x4 dx
= 624.8 units
2. Find the area between the curves y = x and y = x2.
Solution
The required area is represented by the shaded area OBAO as
The points of intersection of the curves, y = x and y = x2, is A (1, 1)
We draw AC perpendicular to x - axis.
Area (OBAO) = Area (ΔOCA) - Area (OCABO) ...(1)
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjAoxYCpHv0sWZSND0FLMr68tPIK752TpilVdq3ZJo6ZcJudhWVhvV1xuSHuX9RfJiM6nMudbU-kg-gUftw4UQIhf9W9xnZUzDAf0jBN5W4RRUIgkmX3ELfP3PaqdBrlopFi3sOp-hQSKuc2clQMex31QfvOntLnC6Bi_QMRnqpk1r_GtPmiSz0uMNi/w120-h95-rw/NCERT%20Solutions%20for%20Chapter%208%20Application%20of%20Integrals%20Class%2012%20Maths%20Misc.%20Exercise%20img%206.JPG)
= 1/2 - 1/3
= 1/6 units
= 1/2 - 1/3
= 1/6 units
3. Find the area of the region lying in the first quadrant and bounded by y = 4x2, x = 0, y = 1 and y = 4.
Solution
The area in the first quadrant bounded by y = 4x2 , x = 0, y = 1 and y = 4 is represented by the shaded area ABCDA as
Area ADCBA = ∫14 y dx
= ∫14 √y/2 dx
= 7/3 units
4. Sketch the graph of y = |x + 3| and evaluate ∫-60 |x + 3| dx.
Solution
The given equation is y = |x + 3|
The corresponding values of x and y are given in the following table :
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgwkbNI1P8DhlM00Rmr04VH_VyPaIMD2Koc3VA_Rc1mfELBYW0vFvCPNqvyS7sxDPdsLwD6xgcHzTOGeKob-fJgLapGQeQ4kuuY19NnnchPrZEdjuK1TNcnkJUuDgGauTwwSiHcSOV1U248Uuq33iUvq6i3wyTbJ2ptnSwU7fFRjgjIdcRA733uGmGR/s1600-rw/NCERT%20Solutions%20for%20Chapter%208%20Application%20of%20Integrals%20Class%2012%20Maths%20Misc.%20Exercise%20img%209.JPG)
On plotting these points, we obtain the graph of y = |x + 3| as follows.![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg0PR8A9igwgRcfzSsRW5CkSff9Y6WQtdXX-exDiL0Qc0vRewn7SN9Rmigoobr6PDBs1lj_h01BJ8F0KSC2k1qDtAWPXMhyLhbZ3_B0dr9SrX1iQrdCj7zAxX07YDLLtToVZf4Ay8VOq7nMoL5JzqS_Z3UgiB9hDMnOtoXI_DV6kwlHJJoC1-xJAz-9/w330-h321-rw/NCERT%20Solutions%20for%20Chapter%208%20Application%20of%20Integrals%20Class%2012%20Maths%20Misc.%20Exercise%20img%2010.JPG)
It is known that, (x + 3) ≤ 0 for -6 ≤ x ≤ -3 and (x + 3) ≥ 0 for -3 ≤ x ≤ 0
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj3jwdzHceN9ZcEQtKn7orIDooNOPlo3YQRafnACNO5tzAPIWrQXibaAvpIc234OeihXEPZx4W661JUUJ-SRtzcTyTxhfyFRxCAgBNWGfv5fnqMJYVVEzyhJGAdjaV_nrB9fXpHvuPFWGwjO_RByRE1BLfcQA8prIsI8C1L4r_WW848m9lfZO1AjI8x/w427-h195-rw/NCERT%20Solutions%20for%20Chapter%208%20Application%20of%20Integrals%20Class%2012%20Maths%20Misc.%20Exercise%20img%2011.JPG)
= 9
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiwcXpul2gdFbbLFpD55uIGFtF3PjvwRay3pxRSe8wW6cu5P5grZ3dsZLasaJ5s1nLGKxtsDMvK_lRXNhBc8c2ysL6-b_u4OKiLPlY3S41PDK23ZU1EW33EvXwlHO1NMU7zXwxh9tknu56P95SuR_9dGFfO84ys7hO26gvotgGn1hj5-duqmqEeDljs/w260-h226-rw/NCERT%20Solutions%20for%20Chapter%208%20Application%20of%20Integrals%20Class%2012%20Maths%20Misc.%20Exercise%20img%2012.JPG)
∴ Required area = Area OABO + Area BCDB
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj_6hPFZZQvCU0R_XtnT37FNfDyB2YQEGG60laEgNntCUO4beqoQNigr1zjV7m2aRUgTvWf8TsboqCbcJLIRa5cGkZZHlPc0DupM9UEX5cUEMZjediBJ6E45hvD_uxUPz3km6vtTiZN4NaRGHZPUlILbwpQGhZ9MGEHlEFQG5cxGklMiAd0JuxHUMOB/w192-h82-rw/NCERT%20Solutions%20for%20Chapter%208%20Application%20of%20Integrals%20Class%2012%20Maths%20Misc.%20Exercise%20img%2013.JPG)
= [ - cosπ + cos 0] + |- cos 2π + cos π|
= 1 + 1+ |(-1-1)|
= 2 + |-2|
= 2 + 2 = 4 units
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiYjjWSeYeAAXYwpNxefiL8NlaI2mgGubkZBnVyjsGNnZdd9eUmzGZPODfjPXJ4NzkoDgUdDsFvGoyYQrVL59hFY0urG2eBs12a0IqDMS26Q0HKXO9MbV5KFC_bwd34tivGAakuFeaxOjAagXwRStISaRvJDcQg2GRgOkGmo0zedlgQBGhSID8QwTZ6/w306-h277-rw/NCERT%20Solutions%20for%20Chapter%208%20Application%20of%20Integrals%20Class%2012%20Maths%20Misc.%20Exercise%20img%2014.JPG)
The points of intersection of both the curves are (0, 0) and (4a/m2, 4a/m)
We draw AC perpendicular to x - axis.![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjpL-JOPN4ndA7bRzMoqw51yNFHGOfHJexcLmYqbD9PVxjIogWqa9d3XT4XhmHpUL3D7xbo96qpsNkgREP3ChoEhL6uC8-wGkdO9A5I_ptoTDZE5ob8zYsZ58BCgq5f4bL_P0ohPII3V3Ui_DEj0bFKqop1jRlOwEmrmwJlvCA0KQCmkyNl07dwUXfV/w220-h396-rw/NCERT%20Solutions%20for%20Chapter%208%20Application%20of%20Integrals%20Class%2012%20Maths%20Misc.%20Exercise%20img%2015.JPG)
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEijix2QclmCJXOtFoMjVEPKuqtwu2aPne-0CS1qFbis8Kyycjbpv6HfClrHMDRjBCMS6hboeQCZMokgCY0vxRLruZPQxUx3ipp3VmxYe6O-vSFZGwEFCr7PHunzjKLQsw3-Gn-xg0gC6fHNFhOY7UBYb-8izk8qB1mLT_tG6m4OIr8x8x4MOvC5nxnX/w298-h425-rw/NCERT%20Solutions%20for%20Chapter%208%20Application%20of%20Integrals%20Class%2012%20Maths%20Misc.%20Exercise%20img%2016.JPG)
The points of intersection of the given curves are A(-2, 3) and (4, 12).
We draw AC and BD perpendicular to x - axis.
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhkaJ5TmK9axp1PAM-4uDl5gG8Tyg87mzdGDg8k7DS_CTTevOcuvIKN1A1dSM-eeVO18eWc_jqV8dTbmVNk03h258feyIS9wJwGYr2EI2RrguEn9IGsfjFJgOo4LHKuGQPoctQDsLKrE0CcHl1AF8W1CEGPVauklTimiV-LYptK1LFZwMIgqOUM-T5o/w274-h272-rw/NCERT%20Solutions%20for%20Chapter%208%20Application%20of%20Integrals%20Class%2012%20Maths%20Misc.%20Exercise%20img%2018.JPG)
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjG4tQGVxKwroZ8Q9H2VPFDP7wb0thM6U0yBSnEQFRLz7f1pe-kQ-yCycaHijPsrBYe1V0dmWpKoYRc7mczFanJ5cv5AMzscNwfw69ETJLfP1XnOiTBxwROJ6zLM8hKZXUnm1Qp6id7480YZuDX2rZLDnBn58NAqvCrWoswfFHpxzLOhi5yDI_WZ1mc/w300-h359-rw/NCERT%20Solutions%20for%20Chapter%208%20Application%20of%20Integrals%20Class%2012%20Maths%20Misc.%20Exercise%20img%2019.JPG)
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEilQHPu7y1lzA8gaKQ0l7ZviaaayPzydU6zxCD_8C0G2Mbp0y77tCxalrLk9XhvjNSh9vYIrdB0OY06jmU5gVco7ibisGrG7_Cq7GHWz06eZWqffm4rIe_5xwqcMXg7saPPy_GbfcIEzTB8sWPQ8wTh-9J7t1opGPgOQrrszio1OujxEn5rzram01M7/w300-h277-rw/NCERT%20Solutions%20for%20Chapter%208%20Application%20of%20Integrals%20Class%2012%20Maths%20Misc.%20Exercise%20img%2020.JPG)
Area BCAB = Area (OBCAO) - Area (OBAO)
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhUkcKk8RYLi_Mc26ywqaSDX5AOlDU5NprFfbPJBF8_wbcK1yvJ8W6wv760mvzDQMKCSIFVeenZcN85gRkuhiyitejR7A1OMhJzvpx_496s72ghnjG1qqgceZc_DUXOmI6erYpiTYE7E_k1uiF6JNI_-czNb7jP7VuFtTqOsceGwNKXg0rUN5BAzYRL/w302-h402-rw/NCERT%20Solutions%20for%20Chapter%208%20Application%20of%20Integrals%20Class%2012%20Maths%20Misc.%20Exercise%20img%2021.JPG)
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgtyg59bFMbFNJJNI0sOYa75d8pka2pBMqeRMjXMzhZTqA1_hwdBIxdwgNtNbCF2_rgZduJx7-dlHeXXVTkm8igNStfc7e5TtUrpBT_65MSZt8GiTlDTkjowRiSV51DmdovQ6JwzoLqWOLThaQZEIn1ASfFTPTHepZJySiF6TEYv7SifSkATv0Et43d/w293-h287-rw/NCERT%20Solutions%20for%20Chapter%208%20Application%20of%20Integrals%20Class%2012%20Maths%20Misc.%20Exercise%20img%2022.JPG)
The point of intersection of the parabola, x2 = y, and the line, y = x + 2, is A(-1, 1).
Area OABCO = Area (BCA) + Area COAC
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjcudl3a9ppmXvpHbm66E8Us8CvD-dlXx-RxMngih5xUnGSwGubbwmeh4bzlTHXhp3yvRA_nCwAKgnf-yBSa5tTfonCy8nCVSsP1R3YwF2VHBWk6wB9XvA6cMtrEUV8pSNlP91DJcOZW6OquLD1t9q2hWxOm1Q2jZsYjZwNrn3FPEkLy1O6hmV2ZB-1/s1600-rw/NCERT%20Solutions%20for%20Chapter%208%20Application%20of%20Integrals%20Class%2012%20Maths%20Misc.%20Exercise%20img%2023.JPG)
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEghACKmQph-kP1Ysa2vET-7-Rj6GaVsjIRNBCRDTs96RlJqRrwP2kSXVDKnxjWLb3CpyCYXb8Eqvgt_NwMJfYrZ5RHZJlHHTN4Ps8_UF17zcZWWV7_AOlZwkusgYHmjZdyDC5HvobJydazZSEUVdnMRmt0T4IABQ4xqY31bUaF6WZSDVirhgEPFB3Zw/w286-h273-rw/NCERT%20Solutions%20for%20Chapter%208%20Application%20of%20Integrals%20Class%2012%20Maths%20Misc.%20Exercise%20img%2024.JPG)
The curve intersects the axes at points A(0, 1), B(1, 0), C(0, -1), and D(-1, 0).
It can be observed that the given curve is symmetrical about x - axis and y - axis.
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh0bNk7FP9rZQsWfXHS8z7vfii-6wfBGEp5yIVx8iHk5XxV7rtuSFNvV8EBZMjc36R5oCSkDI0DVg-OJ5CsjgVB72lp-kyHB3ZY2Gf-uSvy7MauBoMO2vCnHIXiKpBtwHSnFDJe07YnYavuTRejz9ZyU20bMp83Nu0528kZ_4vPTIME1D4IfX-r9_ZU/w293-h280-rw/NCERT%20Solutions%20for%20Chapter%208%20Application%20of%20Integrals%20Class%2012%20Maths%20Misc.%20Exercise%20img%2026.JPG)
It can be observed that the required area is symmetrical about y - axis.
Required area = 2[Area (OCAO) - Area (OCADO)]
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjPUEzCNqBTlR7S16NfRsg4LjB599A8QqHU5Icdx_VqLCwQ-tjQFcn_uAkQ4snW0CVuHtZfy3Orcl4jPLU7pwk1NKxJHh6y_Ro5NoVeQkuJvaygwq7-zChElD4ZdK-0oU5YiFCuA7eFRFU0r1mrSQSL94PRVivACUOXCVsqO92jGWQu5iSi5udpCSCr/w152-h207-rw/NCERT%20Solutions%20for%20Chapter%208%20Application%20of%20Integrals%20Class%2012%20Maths%20Misc.%20Exercise%20img%2027.JPG)
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgQy9zpidN6YQP_uCqVhvtr2Jz02jKiT3drJx1aNs94OIhQxB8fYz6iUSD4aM5-j0ZaZmu0GdZcF8lGFcbQl6328e0uu_reUehk0CpAXeReUpaVk_gAPRolfcN9KJA18V0tBzBhv7tJWtDho5WUCeFViTef32agIS3krxL0gF1t2FLb354syypfE1YB/w321-h312-rw/NCERT%20Solutions%20for%20Chapter%208%20Application%20of%20Integrals%20Class%2012%20Maths%20Misc.%20Exercise%20img%2028.JPG)
Equation of line segment AB is
y - 0 = [(5 - 0)/(4 - 2)] (x - 2)
⇒ 2y = 5x - 10
⇒ y = (5/2) (x - 2) ...(1)
Equation of line segment BC is
y - 5 = [(3 - 5)(6 - 4)] (x - 4)
⇒ 2y - 10 = -2x + 8
⇒ 2y = -2x + 18
⇒ y = -x + 9 ...(2)
Equation of line segment CA is
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjvMzyL8DeSxVrQJvw38-7NrFdpYOGKJyhVcByge7emrzUWFeitoguKgsr5pjCx-yZT5s5eWxJzT2kdo_UIrWywctDTatQ-Iu1T6OCQbN70JGWJUAFHGrOPn0lC3JwFVY7I7NRwmmlbUMD_LsLSxNr5DGtdvy27uTSS4CinfrzkLicirIQjrX0AYdNu/w139-h43-rw/NCERT%20Solutions%20for%20Chapter%208%20Application%20of%20Integrals%20Class%2012%20Maths%20Misc.%20Exercise%20img%2029.JPG)
⇒ -4y + 12 = -3x + 18
⇒ 4y = 3x - 6
⇒ y = (3/4)(x - 2) ...(3)
Area (ΔABC) = Area (ABLA) + Area (BLMCB) - Area(ACMA)
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhozqKJM-9r_ZIWfjBmrmqvLbBH3-_VrVgZHV2qpPYjDTVSXq1pFyzYo89UlVRGz2Pjd7l1a4V8u4bT3UFAMiSxk_nKw6SGwmRFZxEReeQ3SKiUiz-hqigmcPU7HDxP41doukS2eUYLlloQ7Sr9YvE0KfI37GlJl55q-E0A0DnL2szaCIhIMlAKESIY/w408-h204-rw/NCERT%20Solutions%20for%20Chapter%208%20Application%20of%20Integrals%20Class%2012%20Maths%20Misc.%20Exercise%20img%2030.JPG)
= 13 - 6
= 7 units
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhzhirwW2f4jRhnttGDwa3LgTSl-VaUPFVFYBFy6R3enTp-n30v3y3m3GMACSpX243GTRYqE87LX4C4bIoIP6nm3N2ljy4cHX2O7ntp4s4kWp04noqQSVAEz1f1_V2v_E3n5Fv3nUzNkhpBw3WB-i6MOueL4bU5A1E6VCIYcKd2MpQvdsI3furRzo-e/w363-h323-rw/NCERT%20Solutions%20for%20Chapter%208%20Application%20of%20Integrals%20Class%2012%20Maths%20Misc.%20Exercise%20img%2031.JPG)
The area of the region bounded by the lines is the area of ΔABC. AL and CM are the perpendiculars on x - axis.
Area (ΔABC) = Area (ALMCA) - Area (ALB) - Area(CMB)
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh7AilOGRgolkYG3svV7K_90wNojEB2KVZZpy94NgkNeMI7q8VPIB9CH2zkS8-8On0SKTFewix95PYN7xJHdMA4ImWY54VIByUbz_E6Fc5ddxLzNISVfac7zbn2ipFAmMMKFeC-GiypopoEYkuOYxg0a7p6dG1sF7Kmr5dUnqsEuP_KmjB9oejPnloc/w405-h322-rw/NCERT%20Solutions%20for%20Chapter%208%20Application%20of%20Integrals%20Class%2012%20Maths%20Misc.%20Exercise%20img%2032.JPG)
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEigICAgG1qynAvCg5iWKPPFsVQFF1SpuANCU32OEPPWVxJ--OeRxar_zRcjPhUan_U2QUBUCMfqNUP7oBp8WCTW3Lh2qE-Jts0bp0Pt6kUl6ymM6s-qRco8nN4jELvqRpqAE3BieiFZuVwy8lMf_XuNyaS80e4OUghP0kcWRvKXPcPtKyxUIyRyrlHB/w343-h333-rw/NCERT%20Solutions%20for%20Chapter%208%20Application%20of%20Integrals%20Class%2012%20Maths%20Misc.%20Exercise%20img%2033.JPG)
The points of intersection of both the curves are (1/2, √2) and (1/2, -√2).
The required area is given by OABCO.
It can be observed that area OABCO is symmetrical about x - axis.
Area OABCO = 2 × Area OBC
Area OBCO = Area OMC + Area MBC
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjTrCldM1yQunlgae44s1YWPNkKXFiTDcXiPZXyXOj7Qkgdkn-Efb3rt8vbRW9TAhNDFkYz5zXuajRQ2zqbMWIBVbZRSsimljzlj5wNkvaKwcRs2edBBDErTJks4wd9je5Yho41H6hhAnBgoe-ZkSx4mRvsBQgDe3aUnQ-IcOyejVAn-7vcZVQi-jT4/w258-h107-rw/NCERT%20Solutions%20for%20Chapter%208%20Application%20of%20Integrals%20Class%2012%20Maths%20Misc.%20Exercise%20img%2034.JPG)
Put 2x = t![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi5pX6tERRzPxLflipOKjn8cNEGwm97Tc-KGJlFQteRwPT7FVE-_mGW8RaqRlrThc5F831U1o-cH43eH7O6NWEXIqFA9kdjDLkxCe3owHBV-TI_1dX9YX457XIQg7IfbH6KqpdNRwYeh-pM5i1JYEGSPESSmcvevhQbh5oYNU5mY5240ntdsAFWA5kO/w493-h414-rw/NCERT%20Solutions%20for%20Chapter%208%20Application%20of%20Integrals%20Class%2012%20Maths%20Misc.%20Exercise%20img%2035.JPG)
Therefore, the required area is
units
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgWBKd-anOZQpQnddj4Orm4cW4PGkAWkqEAqNV7kE3LVcxj_SoF59Vm6YFub1KIu8AdAUvrbGKEk0kC0kNEdaU_HlstWjtDG0xIOUKrK43y3x5qu79Zu1SK8ble4lM8Qne7SC6jzGonyvUhXSo4lg59Do-ehhcLftw9pVkpOjbixCgyBgFckocGGjIZ/w344-h52-rw/NCERT%20Solutions%20for%20Chapter%208%20Application%20of%20Integrals%20Class%2012%20Maths%20Misc.%20Exercise%20img%2036.JPG)
Required area = A(DCOD) + A(OAB)
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgb9t56z_-p2OYDGdY_ZryBRH7OkuSDypkhtW9eaGGEzvJE1j5KckD4kH3ujBdN-1XY0ZdMYsn4Fud1O6zBEx9jMilA6i6TJUGWVUMba996b3CnuRLwLo4BrjHDQpF31HTfSeJnxHCkiYPExC33t06B1h0STgIb6iQDaudo-g3fDO5kHClb9gWeyeJw/w149-h208-rw/NCERT%20Solutions%20for%20Chapter%208%20Application%20of%20Integrals%20Class%2012%20Maths%20Misc.%20Exercise%20img%2038.JPG)
Since the area is positive
∴ A(DCOD) = 16/4
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhQaVyrS6uzjZLSN08EzeLj36A33ztiMGP04vf3cl41qAiAKlrohBSteCsp4-Liy4Z2I80eCMMFQWlQEERcgD_cdhZllnHkcEdhV8KzTsTsQZnvPF8Xu4ZZY1BV0E6ap4ztGkwR8hgMGdnYWgPPqXk-zshZILz6U3lTSSKqHmjqAeEZ9lubnOMKNxje/w277-h195-rw/NCERT%20Solutions%20for%20Chapter%208%20Application%20of%20Integrals%20Class%2012%20Maths%20Misc.%20Exercise%20img%2039.JPG)
Hence, the required area of the region is 17/4 units
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjR6BSRJiKIhsrw36kBmbpBrfyllVogw7eB3Q9RspLh3KBVbqDE4lCnjs8aebYB3i9IxSLmQK8v0lxP6aNsLebRILSFoOE0NwRA7RCiLUAl7Fe2x-PFoQUeNacnOFyyfo_ljh0pXJvgckcMdmEWza8nl4F9T8mOg8MYbwP31mKjjRrPy3W6zui1ypsr/w258-h270-rw/NCERT%20Solutions%20for%20Chapter%208%20Application%20of%20Integrals%20Class%2012%20Maths%20Misc.%20Exercise%20img%2040.JPG)
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEis2w97P5cBB8HbKlPsUcWULWNg3j6TvS2Nb5nrMNneJuhNfkRgzYBYzdoZQus_34pNq5IGopHMls_xmksohvp_nKfOCoXhMn3FgsrSoqIARpocrFc5tMMeD29rL7-nC51UvmSgKNR9qoUICbq0WsboBESczlOwBevTz9VBro8H_rKFZbokPuKx1xnu/w196-h494-rw/NCERT%20Solutions%20for%20Chapter%208%20Application%20of%20Integrals%20Class%2012%20Maths%20Misc.%20Exercise%20img%2041.JPG)
= 1/3
∴ A(OAB) = 1/3
Thus, the total area = 1/3 + 1/3 = 2/3 units
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg-jGAVp60l8KLKRqIKlUuLVgD5o1b5L1Rc6E8UZphEJpYSiS6Lj0H_qaWnbah7uz7t--K4b2HYtcjfHUgq05ZiP77DpEr-YOXcMnA9Aa18YugPkZAXK1Bv1emlU-D6mSOEp72SVi2O3D2nl-ZWyWhYZdEXCxuSDJnbS7LPiDxDJsc9A0guIp5hOf9R/w116-h155-rw/NCERT%20Solutions%20for%20Chapter%208%20Application%20of%20Integrals%20Class%2012%20Maths%20Misc.%20Exercise%20img%2042.JPG)
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjqtwssqWHCHbpw2YtaPX_ZCE2oduP65vau-yhUXRm7tBiI23oqo1dNyYyASeoBbSi6HkBCuUcLkkN6Cvk8l6fN391jKjZSiVqyTlAQ3zow0kEbRdOPi-FN1rVaP4VDHQhp2ou_vjQTqxHF-2v-PLK8cK-SLghnTXmI_O6Sd9iJjEYkJ6KRu6OHNkww/w278-h274-rw/NCERT%20Solutions%20for%20Chapter%208%20Application%20of%20Integrals%20Class%2012%20Maths%20Misc.%20Exercise%20img%2043.JPG)
Area bounded by the circle and parabola
= 2[Area (OADO) + Area(ADBA)]
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjAmJccbpSSc9MpiGcJagSf6gKAJvzlxugpkhc4jIIu1U2Q8CJQNULxxa6S9pgpWoOlcoWtMXb22XY2lylR4_nzYqGoewUUAkti6NrzBbt8sSMwHN-EF2D5sK3MrujG3NCo-o2p2x4NsudQpfgQirGMCeWXmhBFHltIYnOskhgEZD08cLY7YMZbfY-H/w347-h458-rw/NCERT%20Solutions%20for%20Chapter%208%20Application%20of%20Integrals%20Class%2012%20Maths%20Misc.%20Exercise%20img%2044.JPG)
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgka_EBTtrkT7CGCWst4ausN7jGvLkf4gu0SYOLhPHj3WsYMUoBH582jnuAYFEgO1HaD9nVcb_yUEa1rIMMliKyOwxsp41CDANbUr3YA3R-v6XCZBLS4ai0iTvn4DT3gV_KKtDrX2ngt-1Sv9yEBuYsxVYLUX99RVSggrYkrQiuBJGrjJ4hYhpfjupS/w249-h119-rw/NCERT%20Solutions%20for%20Chapter%208%20Application%20of%20Integrals%20Class%2012%20Maths%20Misc.%20Exercise%20img%2045.JPG)
Thus, the correct answer is C.
The corresponding values of x and y are given in the following table :
On plotting these points, we obtain the graph of y = |x + 3| as follows.
It is known that, (x + 3) ≤ 0 for -6 ≤ x ≤ -3 and (x + 3) ≥ 0 for -3 ≤ x ≤ 0
= 9
5. Find the area bounded by the curve y = sin x between x = 0 and x = 2Ï€.
Solution
The graph of y = sin x can be drawn as
∴ Required area = Area OABO + Area BCDB
= [ - cosπ + cos 0] + |- cos 2π + cos π|
= 1 + 1+ |(-1-1)|
= 2 + |-2|
= 2 + 2 = 4 units
6. Find the area enclosed between the parabola y2 = 4ax and the line y = mx.
Solution
The area enclosed between the parabola, y2 = 4ax and the line y = mx, is represented by the shaded area OABO as
The points of intersection of both the curves are (0, 0) and (4a/m2, 4a/m)
We draw AC perpendicular to x - axis.
Area OABO = Area OCABO - Area (ΔOCA)
7. Find the area enclosed by the parabola 4y = 3x2 and the line 2y = 3x + 12
Solution
The area enclosed between the parabola, 4y = 3x2 , and the line, 2y = 3x + 12, is represented by the shaded area OBAO as
The points of intersection of the given curves are A(-2, 3) and (4, 12).
We draw AC and BD perpendicular to x - axis.
Area OBAO = Area CDBA - (Area ODBO + Area OACO)
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi86MRJzCFzlrCPePIcNO_Sixqie8-NKhnFVxS91dtbl4wJrcO1LiOuOMknybVJ7z8JtlCnyz3WcykdU3558joAVnS9-slsNkqC_x-Ft5cf-rUvITqBkkb1KHF-KulJZkY7K5mlCpnZkpYN1KGzfo9ikOj-WTzOl1P4QdsjihiMUMFL2L-s7O8cbqng/w226-h194-rw/NCERT%20Solutions%20for%20Chapter%208%20Application%20of%20Integrals%20Class%2012%20Maths%20Misc.%20Exercise%20img%2017.JPG)
= 45 - 18
= 27 units
= 45 - 18
= 27 units
8. Find the area of the smaller region bounded by the ellipse x2 /9 + y2 /4 and the line x/3 + y/2 = 1.
Solution
The area of the smaller region bounded by the ellipse, x2/9 + y2/4 = 1, and the line x/3 + y/2 = 1 , is represented by the shaded region BCAB as
Area BCAB = Area (OBCAO) - Area (OBAO)
9. Find the area of the smaller region bounded by the ellipse x2 /a2 + y2 /b2 = 1 and the line x/a + y/b = 1.
Solution
The area of the smaller region bounded by the ellipse, x2 /a2 + y2 /b2 = 1 and the line, x/a + y/b = 1, is represented by the the shaded region BCAB as
Area BCAB = Area (OBCAO) - Area (OBAO)
10. Find the area of the region enclosed by the parabola x2 = y, the line y = x + 2 and x-axis.
Solution
The area of the region enclosed by the parabola, x2 = y , the line, y = x + 2 and x - axis is represented by the shaded region OABCO as
The point of intersection of the parabola, x2 = y, and the line, y = x + 2, is A(-1, 1).
Area OABCO = Area (BCA) + Area COAC
11. Using the method of integration find the area bounded by the curve |x| + |y| = 1.
[Hint: The required region is bounded by lines x + y = 1, x– y = 1, – x + y = 1 and – x – y = 1].
[Hint: The required region is bounded by lines x + y = 1, x– y = 1, – x + y = 1 and – x – y = 1].
Solution
The area bounded by the curve, |x| + |y| = 1, is represented by the shaded region ADCB as
The curve intersects the axes at points A(0, 1), B(1, 0), C(0, -1), and D(-1, 0).
It can be observed that the given curve is symmetrical about x - axis and y - axis.
∴Area ADCB = 4 × Area OBAO
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjMe771iI6vE9N8xG3ks5xFekqT0wHzJ-9BCAoyZtkw8Razp7M_7wZyWBAja7QoJtvH0hKFoBG7YiraB8P0Hu5Wdim4BE9uiMcnGEWsrKnJbbpqfsY13VXgAU5BdvblMo7yt8ATBYfrU047jgwSW9yoZzac1d0lPQSNF8yMmOm0tTjdrnornhiIxX2E/w99-h150-rw/NCERT%20Solutions%20for%20Chapter%208%20Application%20of%20Integrals%20Class%2012%20Maths%20Misc.%20Exercise%20img%2025.JPG)
= 4(1/2)
= 2 units.
= 4(1/2)
= 2 units.
12. Find the area bounded by curves {(x, y) : y ≥ x2 and y = |x|}.
Solution
The area bounded by the curves, {(x, y) : y ≥ x2 and y = |x|}, is represented by the shaded region as
It can be observed that the required area is symmetrical about y - axis.
Required area = 2[Area (OCAO) - Area (OCADO)]
13. Using the method of integration find the area of the triangle ABC, coordinates of whose vertices are A(2, 0), B (4, 5) and C (6, 3).
Solution
The Vertices of ΔABC are A(2, 0), B(4, 5) and C(6, 3).
Equation of line segment AB is
y - 0 = [(5 - 0)/(4 - 2)] (x - 2)
⇒ 2y = 5x - 10
⇒ y = (5/2) (x - 2) ...(1)
Equation of line segment BC is
y - 5 = [(3 - 5)(6 - 4)] (x - 4)
⇒ 2y - 10 = -2x + 8
⇒ 2y = -2x + 18
⇒ y = -x + 9 ...(2)
Equation of line segment CA is
⇒ -4y + 12 = -3x + 18
⇒ 4y = 3x - 6
⇒ y = (3/4)(x - 2) ...(3)
Area (ΔABC) = Area (ABLA) + Area (BLMCB) - Area(ACMA)
= 13 - 6
= 7 units
14. Using the method of integration find the area of the region bounded by lines: 2x + y = 4, 3x – 2y = 6 and x – 3y + 5 = 0
Solution
The given equations of lines are
2x + y= 4 ...(1)
3x -2 y = 6 ...(2)
And, x - 3y + 5 = 0 ...(3)
2x + y= 4 ...(1)
3x -2 y = 6 ...(2)
And, x - 3y + 5 = 0 ...(3)
The area of the region bounded by the lines is the area of ΔABC. AL and CM are the perpendiculars on x - axis.
Area (ΔABC) = Area (ALMCA) - Area (ALB) - Area(CMB)
15. Find the area of the region {(x, y) : y2 ≤ 4x, 4x2 + 4y2 ≤ 9}
Solution
The area bounded by the curves, {(x, y) : y2 ≤ 4x, 4x2 + 4y2 ≤ 9}, is represented as
The points of intersection of both the curves are (1/2, √2) and (1/2, -√2).
The required area is given by OABCO.
It can be observed that area OABCO is symmetrical about x - axis.
Area OABCO = 2 × Area OBC
Area OBCO = Area OMC + Area MBC
Put 2x = t
⇒ dx = dt/2
When x = 3/2, t = 3 and when x = 1/2, t = 1
When x = 3/2, t = 3 and when x = 1/2, t = 1
Therefore, the required area is
16. Choose the correct answer Area bounded by the curve y = x3, the x-axis and the ordinates x = –2 and x = 1 is
A. – 9
B. -15/4
C. 15/4
D. 17/4
A. – 9
B. -15/4
C. 15/4
D. 17/4
Solution
Required area = A(DCOD) + A(OAB)
Since the area is positive
∴ A(DCOD) = 16/4
Hence, the required area of the region is 17/4 units
17. Choose the correct answer The area bounded by the curve y = x | x| ,, x-axis and the ordinates x = –1 and x = 1 is given by
[Hint: y = x2 if x > 0 and y = –x2 if x < 0]
A. 0
B. 1/3
C. 2/3
D. 4/3
[Hint: y = x2 if x > 0 and y = –x2 if x < 0]
A. 0
B. 1/3
C. 2/3
D. 4/3
Solution
Required area = A(DCOD) + A(OAB)
= 1/3
∴ A(OAB) = 1/3
Thus, the total area = 1/3 + 1/3 = 2/3 units
Hence, the required area of the region is 2/3 units.
18. Choose the correct answer The area of the circle x2 + y2 = 16 exterior to the parabola y2 = 6x is
Solution
The given equations are
x2 + y2 = 16 ...(1)
y2 = 6x ...(2)
x2 + y2 = 16 ...(1)
y2 = 6x ...(2)
Area bounded by the circle and parabola
= 2[Area (OADO) + Area(ADBA)]
Area of circle = π(r)2
= π(4)2 = 16π units
= π(4)2 = 16π units
Thus, the correct answer is C.
19. The area bounded by the y-axis, y = cos x and y = sin x when 0 <= x <= π2
(A) 2 ( 2 −1)
(B) √2-1
(C) √2+1
(A) 2 ( 2 −1)
(B) √2-1
(C) √2+1
(D) √2
Solution
The given equation are
y = cos x ...(1)
And, y = sin x ...(2)
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgxrl5KYoXhkfSXXeOx1OUhA_aYa56dGdbETkT9W8w1InfVZhGteDBMcedg72JBDy1stX7-c8rxlC-HUB7OzmvrKkwu94-zQvUc-6WpVELnxKkKxYwdxJVgwo-RazaZpl4iogO75b2uf-BQ1tq3wZePPuYtsXujqVW1RdwjE-q5vAbX4c_Z_5_tczca/w298-h267-rw/NCERT%20Solutions%20for%20Chapter%208%20Application%20of%20Integrals%20Class%2012%20Maths%20Misc.%20Exercise%20img%2046.JPG)
Required area = Area (ABLA) + area (OBLO)![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjeU2kc73IIRd3pGqcNmayX-JRcSio2BwcBiBNUmDZGVTTiHMJUaXedua6nFz2rnDivLr3FVTbm2EomR3vgcY8EqdlVSX_M7i0DlWe4G8QjtjZSKmjjOIFwqJBPUEOJKLHCmlJpZazFeIQQu3GVtMsxKlNikLw4LyDC2-_s-aMxRDElsze96TYVSUeH/w220-h116-rw/NCERT%20Solutions%20for%20Chapter%208%20Application%20of%20Integrals%20Class%2012%20Maths%20Misc.%20Exercise%20img%2047.JPG)
y = cos x ...(1)
And, y = sin x ...(2)
Required area = Area (ABLA) + area (OBLO)
Integrating by parts, we obtain
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiXmBlI-AQPWNptn7owsVOs_dzoAf2DveupvKWBiw7R7fsixSRtnIVuyi7si8q10PGFneFjbgRUdga3h7LJGGHW-b6HM41CiP27jQqxaN5Wz1HYQJpmVMuUtLDrDGf02T0aLmOaQsbl7Rel0111O9tpPO_9EJ-ngcZ076zTT0AsuO4BVVlWjCVeYlsi/w471-h206-rw/NCERT%20Solutions%20for%20Chapter%208%20Application%20of%20Integrals%20Class%2012%20Maths%20Misc.%20Exercise%20img%2048.JPG)
= √2 - 1 units
= √2 - 1 units
Thus, the correct answer is B.