10:[["$","title","0",{"children":"CUET Mathematics - The value of the definite integral I = _-1^1 11 + sqrt(e^x) dx is: | PYQs + Solutions | AfterBoards"}],["$","meta","1",{"name":"description","content":"undefined The integral to evaluate is: I = _-1^1 11 + sqrt(e^x) dx This integral has symmetric limits from -1 to 1. For such integrals, the following property can be used: I = _-1^1 f(x) dx = _-1^1 f(-x) dx

Given f(x) = 11 + sqrt(e^x), substitute -x: f(-x) = 11 + e^-x Since e^-x = 1/e^x: f(-x) = 11 + 1sqrt(e^x) Multiplying numerator and denominator by sqrt(e^x): f(-x) = sqrt(e^x)sqrt(e^x) + 1 f(-x) = sqrt(e^x)1 + sqrt(e^x)

Adding f(x) and f(-x): f(x) + f(-x) = 11 + sqrt(e^x) + sqrt(e^x)1 + sqrt(e^x) f(x) + f(-x) = 1 + sqrt(e^x)1 + sqrt(e^x) f(x) + f(-x) = 1

Since I = _-1^1 f(x) dx and I = _-1^1 f(-x) dx: 2I = _-1^1 f(x) dx + _-1^1 f(-x) dx 2I = _-1^1 [f(x) + f(-x)] dx 2I = _-1^1 1 dx 2I = [x]_-1^1 2I = 1 - (-1) 2I = 2

Therefore: I = 1"}],["$","link","2",{"rel":"manifest","href":"https://www.afterboards.in/manifest.json","crossOrigin":"$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"}],["$","meta","7",{"property":"og:title","content":"CUET Mathematics - The value of the definite integral I = _-1^1 11 + sqrt(e^x) dx is: | PYQs + Solutions"}],["$","meta","8",{"property":"og:description","content":"undefined The integral to evaluate is: I = _-1^1 11 + sqrt(e^x) dx This integral has symmetric limits from -1 to 1. For such integrals, the following property can be used: I = _-1^1 f(x) dx = _-1^1 f(-x) dx

Adding f(x) and f(-x): f(x) + f(-x) = 11 + sqrt(e^x) + sqrt(e^x)1 + sqrt(e^x) f(x) + f(-x) = 1 + sqrt(e^x)1 + sqrt(e^x) f(x) + f(-x) = 1

Since I = _-1^1 f(x) dx and I = _-1^1 f(-x) dx: 2I = _-1^1 f(x) dx + _-1^1 f(-x) dx 2I = _-1^1 [f(x) + f(-x)] dx 2I = _-1^1 1 dx 2I = [x]_-1^1 2I = 1 - (-1) 2I = 2

Therefore: I = 1"}],["$","meta","9",{"property":"og:url","content":"https://www.afterboards.in/past-year-questions/cuet-mathematics/2025/slot-6/56"}],["$","meta","10",{"property":"og:site_name","content":"AfterBoards | CUET Mathematics 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"}],["$","meta","15",{"name":"twitter:title","content":"CUET Mathematics - The value of the definite integral I = _-1^1 11 + sqrt(e^x) dx is: | PYQs + Solutions"}],["$","meta","16",{"name":"twitter:description","content":"undefined The integral to evaluate is: I = _-1^1 11 + sqrt(e^x) dx This integral has symmetric limits from -1 to 1. For such integrals, the following property can be used: I = _-1^1 f(x) dx = _-1^1 f(-x) dx

Adding f(x) and f(-x): f(x) + f(-x) = 11 + sqrt(e^x) + sqrt(e^x)1 + sqrt(e^x) f(x) + f(-x) = 1 + sqrt(e^x)1 + sqrt(e^x) f(x) + f(-x) = 1

Since I = _-1^1 f(x) dx and I = _-1^1 f(-x) dx: 2I = _-1^1 f(x) dx + _-1^1 f(-x) dx 2I = _-1^1 [f(x) + f(-x)] dx 2I = _-1^1 1 dx 2I = [x]_-1^1 2I = 1 - (-1) 2I = 2

Therefore: I = 1"}],"$L17","$L18","$L19","$L1a","$L1b","$L1c","$L1d","$L1e","$L1f","$L20"]

CUET Mathematics 2025 15 May Shift 2 PYQs

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