9:{"metadata":[["$","title","0",{"children":"2024 Slot 1 - x^n+1-x d x= | PYQs + Solutions | AfterBoards"}],["$","meta","1",{"name":"description","content":"undefined We need to evaluate x^n+1-x dx

First, <=t's rewrite the denominator: x^n+1 - x = x(x^n - 1) So our integral becomes: x(x^n-1) dx

Using partial fractions: x(x^n-1) = Ax + Bx^n-1 Multiplying all terms by x(x^n-1): = A(x^n-1) + Bx When x = 0: = A(-1), so A = - When x = 1: = B, so B = Therefore: x^n+1-x = -x + x^n-1

Integrating each term: x^n+1-xdx = -xdx + x^n-1dx = -|x| + 1x^n-1dx

For the second integral, use substitution t = x^n: dt = nx^n-1dx → dx = dtnx^n-1 1x^n-1dx = 1t-1 dtnx^n-1 Since x^n-1 = tx, we have: 1x^n-1dx = 1n xt(t-1)dt

Let's try direct substitution with t = x^n: x(x^n-1)dx = n dtt(t-1) = n (1t-1 - 1t)dt = n[|t-1| - |t|] + C = n|t-1t| + C

Substituting back t = x^n: x^n+1-xdx=n|x^n-1x^n| + 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"}],["$","meta","7",{"property":"og:title","content":"2024 Slot 1 - x^n+1-x d x= | PYQs + Solutions"}],["$","meta","8",{"property":"og:description","content":"undefined We need to evaluate x^n+1-x dx

First, <=t's rewrite the denominator: x^n+1 - x = x(x^n - 1) So our integral becomes: x(x^n-1) dx

Using partial fractions: x(x^n-1) = Ax + Bx^n-1 Multiplying all terms by x(x^n-1): = A(x^n-1) + Bx When x = 0: = A(-1), so A = - When x = 1: = B, so B = Therefore: x^n+1-x = -x + x^n-1

Integrating each term: x^n+1-xdx = -xdx + x^n-1dx = -|x| + 1x^n-1dx

For the second integral, use substitution t = x^n: dt = nx^n-1dx → dx = dtnx^n-1 1x^n-1dx = 1t-1 dtnx^n-1 Since x^n-1 = tx, we have: 1x^n-1dx = 1n xt(t-1)dt

Let's try direct substitution with t = x^n: x(x^n-1)dx = n dtt(t-1) = n (1t-1 - 1t)dt = n[|t-1| - |t|] + C = n|t-1t| + C

Substituting back t = x^n: x^n+1-xdx=n|x^n-1x^n| + C"}],["$","meta","9",{"property":"og:url","content":"https://www.afterboards.in/cuet-past-year-questions/cuet-mat-2024-01/6"}],["$","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"}],["$","meta","15",{"name":"twitter:title","content":"2024 Slot 1 - x^n+1-x d x= | PYQs + Solutions"}],["$","meta","16",{"name":"twitter:description","content":"undefined We need to evaluate x^n+1-x dx

First, <=t's rewrite the denominator: x^n+1 - x = x(x^n - 1) So our integral becomes: x(x^n-1) dx

Using partial fractions: x(x^n-1) = Ax + Bx^n-1 Multiplying all terms by x(x^n-1): = A(x^n-1) + Bx When x = 0: = A(-1), so A = - When x = 1: = B, so B = Therefore: x^n+1-x = -x + x^n-1

Integrating each term: x^n+1-xdx = -xdx + x^n-1dx = -|x| + 1x^n-1dx

For the second integral, use substitution t = x^n: dt = nx^n-1dx → dx = dtnx^n-1 1x^n-1dx = 1t-1 dtnx^n-1 Since x^n-1 = tx, we have: 1x^n-1dx = 1n xt(t-1)dt

Let's try direct substitution with t = x^n: x(x^n-1)dx = n dtt(t-1) = n (1t-1 - 1t)dt = n[|t-1| - |t|] + C = n|t-1t| + C

Substituting back t = x^n: x^n+1-xdx=n|x^n-1x^n| + C"}],"$L18","$L19","$L1a","$L1b","$L1c","$L1d","$L1e","$L1f","$L20","$L21"],"error":null,"digest":"$undefined"}

2024 Slot 1 PYQs

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