10:[["$","title","0",{"children":"CUET Mathematics - e^-x( x + ^2 x)dx = | PYQs + Solutions | AfterBoards"}],["$","meta","1",{"name":"description","content":"undefined The product ru<= states: d/dx[f(x) g(x)] = f'(x) g(x) + f(x) g'(x) Key derivatives to recall: d/dx(e^-x) = -e^-x d/dx( x) = -^2 x

Consider the function e^-x x and apply the product ru<=: d/dx[e^-x x] = d/dx(e^-x) x + e^-x d/dx( x) = (-e^-x) x + e^-x (-^2 x) = -e^-x x - e^-x^2 x = -e^-x( x + ^2 x)

The derivative gives -e^-x( x + ^2 x), which differs from the integrand by a negative sign. Therefore: d/dx[-e^-x x] = e^-x( x + ^2 x)

e^-x( x + ^2 x)dx = -e^-x x + c"}],["$","link","2",{"rel":"manifest","href":"https://www.afterboards.in/manifest.json","crossOrigin":"$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"}],["$","meta","7",{"property":"og:title","content":"CUET Mathematics - e^-x( x + ^2 x)dx = | PYQs + Solutions"}],["$","meta","8",{"property":"og:description","content":"undefined The product ru<= states: d/dx[f(x) g(x)] = f'(x) g(x) + f(x) g'(x) Key derivatives to recall: d/dx(e^-x) = -e^-x d/dx( x) = -^2 x

Consider the function e^-x x and apply the product ru<=: d/dx[e^-x x] = d/dx(e^-x) x + e^-x d/dx( x) = (-e^-x) x + e^-x (-^2 x) = -e^-x x - e^-x^2 x = -e^-x( x + ^2 x)

The derivative gives -e^-x( x + ^2 x), which differs from the integrand by a negative sign. Therefore: d/dx[-e^-x x] = e^-x( x + ^2 x)

e^-x( x + ^2 x)dx = -e^-x x + c"}],["$","meta","9",{"property":"og:url","content":"https://www.afterboards.in/past-year-questions/cuet-mathematics/2025/slot-14/34"}],["$","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"}],["$","meta","15",{"name":"twitter:title","content":"CUET Mathematics - e^-x( x + ^2 x)dx = | PYQs + Solutions"}],["$","meta","16",{"name":"twitter:description","content":"undefined The product ru<= states: d/dx[f(x) g(x)] = f'(x) g(x) + f(x) g'(x) Key derivatives to recall: d/dx(e^-x) = -e^-x d/dx( x) = -^2 x

Consider the function e^-x x and apply the product ru<=: d/dx[e^-x x] = d/dx(e^-x) x + e^-x d/dx( x) = (-e^-x) x + e^-x (-^2 x) = -e^-x x - e^-x^2 x = -e^-x( x + ^2 x)

The derivative gives -e^-x( x + ^2 x), which differs from the integrand by a negative sign. Therefore: d/dx[-e^-x x] = e^-x( x + ^2 x)

e^-x( x + ^2 x)dx = -e^-x x + c"}],["$","meta","17",{"name":"twitter:image","content":"https://www.afterboards.in/pyp/AfterBoards%20CUET%20Past%20Year%20Papers%20with%20Solutions.webp"}],["$","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"}],["$","link","21",{"rel":"icon","href":"https://www.afterboards.in/icons/android/android-launchericon-512-512.png","sizes":"512x512","type":"image/png"}],["$","link","22",{"rel":"icon","href":"https://www.afterboards.in/icons/android/android-launchericon-192-192.png","sizes":"192x192","type":"image/png"}],"$L17","$L18","$L19","$L1a"]

CUET Mathematics 2025 27 May Shift 1 PYQs

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