9:{"metadata":[["$","title","0",{"children":"CUET Mathematics 2024 - The value of the integral _ _e 2^ _e 3 e^2 x-1e^2 x+1 d x is : | PYQs + Solutions | AfterBoards"}],["$","meta","1",{"name":"description","content":"undefined Simplify: e^2x-1e^2x+1 = e^2x+1-2e^2x+1 = 1 - 2e^2x+1 So our integral becomes: __e 2^_e 3 (1 - 2e^2x+1) dx = __e 2^_e 3 dx - 2__e 2^_e 3 1e^2x+1 dx

The first part is straightforward: __e 2^_e 3 dx = x|__e 2^_e 3 = _e 3 - _e 2 = _e 32

For the second part, we use substitution u = e^2x, giving: du = 2e^2xdx and dx = du2u When x = _e 2, u = e^2_e 2 = 4 When x = _e 3, u = e^2_e 3 = 9 The integral becomes: _4^9 1u+1 du2u = 12_4^9 1u(u+1) du

Using partial fractions: 1u(u+1) = 1u - 1u+1 Therefore: 12_4^9 1u(u+1) du = 12_4^9 (1u - 1u+1) du = 12[|u| - |u+1|]_4^9 = 12[|uu+1|]_4^9 = 12[|910| - |45|] = 12|9/104/5| = 12|9 510 4| = 12|4540| = 12|98|

Combining the results: __e 2^_e 3 e^2x-1e^2x+1 dx = _e 32 - 98 = 32 - 98 = (3/29/8) = (3 82 9) = (2418) = (43) The value of the integral is (43) = _e 4- _e 3"}],["$","link","2",{"rel":"manifest","href":"https://www.afterboards.in/manifest.json","crossOrigin":"$undefined"}],["$","meta","3",{"name":"keywords","content":"CUETMAT,MAT,Calculus,Integrals"}],["$","meta","4",{"name":"robots","content":"index, follow"}],["$","meta","5",{"name":"category","content":"education"}],["$","meta","6",{"name":"google-site-verification","content":"google"}],["$","meta","7",{"property":"og:title","content":"CUET Mathematics 2024 - The value of the integral _ _e 2^ _e 3 e^2 x-1e^2 x+1 d x is : | PYQs + Solutions"}],["$","meta","8",{"property":"og:description","content":"undefined Simplify: e^2x-1e^2x+1 = e^2x+1-2e^2x+1 = 1 - 2e^2x+1 So our integral becomes: __e 2^_e 3 (1 - 2e^2x+1) dx = __e 2^_e 3 dx - 2__e 2^_e 3 1e^2x+1 dx

The first part is straightforward: __e 2^_e 3 dx = x|__e 2^_e 3 = _e 3 - _e 2 = _e 32

Combining the results: __e 2^_e 3 e^2x-1e^2x+1 dx = _e 32 - 98 = 32 - 98 = (3/29/8) = (3 82 9) = (2418) = (43) The value of the integral is (43) = _e 4- _e 3"}],["$","meta","9",{"property":"og:url","content":"https://www.afterboards.in/cuet-past-year-questions/cuet-mat-2024-01/19"}],["$","meta","10",{"property":"og:site_name","content":"AfterBoards | CUET Preparation"}],["$","meta","11",{"property":"og:image","content":"https://www.afterboards.in/pyp/AfterBoards%20CUET%20Past%20Year%20Papers%20with%20Solutions.webp"}],["$","meta","12",{"property":"og:type","content":"website"}],["$","meta","13",{"name":"twitter:card","content":"summary_large_image"}],["$","meta","14",{"name":"twitter:site","content":"afterboards.in"}],["$","meta","15",{"name":"twitter:title","content":"CUET Mathematics 2024 - The value of the integral _ _e 2^ _e 3 e^2 x-1e^2 x+1 d x is : | PYQs + Solutions"}],["$","meta","16",{"name":"twitter:description","content":"undefined Simplify: e^2x-1e^2x+1 = e^2x+1-2e^2x+1 = 1 - 2e^2x+1 So our integral becomes: __e 2^_e 3 (1 - 2e^2x+1) dx = __e 2^_e 3 dx - 2__e 2^_e 3 1e^2x+1 dx

The first part is straightforward: __e 2^_e 3 dx = x|__e 2^_e 3 = _e 3 - _e 2 = _e 32

Combining the results: __e 2^_e 3 e^2x-1e^2x+1 dx = _e 32 - 98 = 32 - 98 = (3/29/8) = (3 82 9) = (2418) = (43) The value of the integral is (43) = _e 4- _e 3"}],"$L18","$L19","$L1a","$L1b","$L1c","$L1d","$L1e","$L1f","$L20","$L21"],"error":null,"digest":"$undefined"}

CUET Mathematics 2024 PYQs

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