10:[["$","title","0",{"children":"CUET Mathematics - _0^a sqrt(x)sqrt(x) + sqrt(a-x) dx is equal to | PYQs + Solutions | AfterBoards"}],["$","meta","1",{"name":"description","content":"undefined Let I = _0^a sqrt(x)sqrt(x) + sqrt(a-x) dx ... (1)

Using the property _0^a f(x)dx = _0^a f(a-x)dx, replace x with (a-x): I = _0^a sqrt(a-x)sqrt(a-x) + sqrt(a-(a-x)) dx Since a-(a-x) = x: I = _0^a sqrt(a-x)sqrt(a-x) + sqrt(x) dx ... (2)

Adding equations (1) and (2): I + I = _0^a sqrt(x)sqrt(x) + sqrt(a-x) dx + _0^a sqrt(a-x)sqrt(x) + sqrt(a-x) dx 2I = _0^a [sqrt(x) + sqrt(a-x)sqrt(x) + sqrt(a-x)] dx 2I = _0^a 1 dx

2I = [x]_0^a 2I = a - 0 2I = a I = a2

Therefore, _0^a sqrt(x)sqrt(x) + sqrt(a-x) dx = a2"}],["$","link","2",{"rel":"manifest","href":"https://www.afterboards.in/manifest.json","crossOrigin":"$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"}],["$","meta","7",{"property":"og:title","content":"CUET Mathematics - _0^a sqrt(x)sqrt(x) + sqrt(a-x) dx is equal to | PYQs + Solutions"}],["$","meta","8",{"property":"og:description","content":"undefined Let I = _0^a sqrt(x)sqrt(x) + sqrt(a-x) dx ... (1)

Using the property _0^a f(x)dx = _0^a f(a-x)dx, replace x with (a-x): I = _0^a sqrt(a-x)sqrt(a-x) + sqrt(a-(a-x)) dx Since a-(a-x) = x: I = _0^a sqrt(a-x)sqrt(a-x) + sqrt(x) dx ... (2)

Adding equations (1) and (2): I + I = _0^a sqrt(x)sqrt(x) + sqrt(a-x) dx + _0^a sqrt(a-x)sqrt(x) + sqrt(a-x) dx 2I = _0^a [sqrt(x) + sqrt(a-x)sqrt(x) + sqrt(a-x)] dx 2I = _0^a 1 dx

2I = [x]_0^a 2I = a - 0 2I = a I = a2

Therefore, _0^a sqrt(x)sqrt(x) + sqrt(a-x) dx = a2"}],["$","meta","9",{"property":"og:url","content":"https://www.afterboards.in/past-year-questions/cuet-mathematics/2025/slot-20/6"}],["$","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"}],["$","meta","15",{"name":"twitter:title","content":"CUET Mathematics - _0^a sqrt(x)sqrt(x) + sqrt(a-x) dx is equal to | PYQs + Solutions"}],["$","meta","16",{"name":"twitter:description","content":"undefined Let I = _0^a sqrt(x)sqrt(x) + sqrt(a-x) dx ... (1)

Using the property _0^a f(x)dx = _0^a f(a-x)dx, replace x with (a-x): I = _0^a sqrt(a-x)sqrt(a-x) + sqrt(a-(a-x)) dx Since a-(a-x) = x: I = _0^a sqrt(a-x)sqrt(a-x) + sqrt(x) dx ... (2)

Adding equations (1) and (2): I + I = _0^a sqrt(x)sqrt(x) + sqrt(a-x) dx + _0^a sqrt(a-x)sqrt(x) + sqrt(a-x) dx 2I = _0^a [sqrt(x) + sqrt(a-x)sqrt(x) + sqrt(a-x)] dx 2I = _0^a 1 dx

2I = [x]_0^a 2I = a - 0 2I = a I = a2

Therefore, _0^a sqrt(x)sqrt(x) + sqrt(a-x) dx = a2"}],["$","meta","17",{"name":"twitter:image","content":"https://www.afterboards.in/pyp/AfterBoards%20CUET%20Past%20Year%20Papers%20with%20Solutions.webp"}],["$","link","18",{"rel":"shortcut icon","href":"https://www.afterboards.in/favicon.ico","type":"image/x-icon","sizes":"48x48"}],["$","link","19",{"rel":"icon","href":"https://www.afterboards.in/favicon.ico","type":"image/x-icon","sizes":"48x48"}],["$","link","20",{"rel":"icon","href":"https://www.afterboards.in/icons/windows11/Square44x44Logo.altform-lightunplated_targetsize-256.png","sizes":"256x256","type":"image/png"}],"$L17","$L18","$L19","$L1a","$L1b","$L1c"]

CUET Mathematics 2025 3 June Shift 2 PYQs

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