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Solve a linear equation, then integrate the solution or find an area

This type of question, asked 7 times: 2021 Paper 2, 27 Jun 2022, 28 Jun 2022, 26 Jul 2022, 5 Apr 2024, 22 Jan 2025, 7 Apr 2025. All 7 questions below, oldest first.

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  1. Question 1

    2021 Paper 2

    78
    If a curve y=f(x)y=f(x) passes through the point (1,2)(1,2) and satisfies xdy dx+y=bx4x\frac{dy}{\text{ }dx}+y= bx^{4}, then for what value of b,∫12f(x)dx=625?b,\int_{1}^{2} f(x)dx=\frac{62}{5}?

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  2. Question 2

    27 Jun 2022

    178
    Let y=y(x)y = y(x) be the solution of the differential equation (1−x2)dy=(xy+(x3+2)1−x2)dx,−1<x<1\left( 1 -x^{2} \right)dy=\left( xy+\left( x^{3}+ 2 \right)\sqrt{1 -x^{2}} \right)dx, - 1 <x< 1and y(0)=0y(0) = 0 if ∫−12121−x2y(x)dx=k\int_{-\frac{1}{2}}^{\frac{1}{2}} \sqrt{1 -x^{2}}y(x)dx=k then k−1k^{- 1} is equal to:

  3. Question 3

    28 Jun 2022

    161
    Let the slope of the tangent to a curve y=f(x)y = f(x) at (x,y)(x,y) be given by 2tan⁡x(cos⁡x−y)2\tan x(\cos x - y). if the curve passes through the point (π/4,0)(\pi/4,0), then the value of ∫0π/2ydx\int_{0}^{\pi/2} ydx is equal to

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  4. Question 4

    26 Jul 2022

    160
    Let the solution curve y=f(x)y = f(x) of the differential equation dydx+xyx2−1=x4+2x1−x2,x∈(−1,1)\frac{dy}{dx}+\frac{xy}{x^{2}- 1}=\frac{x^{4}+ 2x}{\sqrt{1 -x^{2}}},x \in ( - 1,1) pass through the origin. Then ∫−3232f(x)dx\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x)dx is equal to

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  5. Question 5

    5 Apr 2024

    172
    Let y=y(x)y = y(x) be the solution of the differential equation dydx+2x(1+x2)2y=xe1(1+x2);y(0)=0\frac{dy}{dx}+\frac{2x}{\left( 1 +x^{2} \right)^{2}}y = xe^{\frac{1}{\left( 1 +x^{2} \right)}};y(0) = 0. Then the area enclosed by the curve f(x)=y(x)e−1(1+x2)f(x) = y(x)e^{-\frac{1}{\left( 1 +x^{2} \right)}} and the line y−x=4y - x = 4 is

  6. Question 6

    22 Jan 2025

    146
    Let y=f(x)y=f(x) be the solution of the differential equation dydx+xyx2−1=x6+4x1−x2, −1<x<1\frac{dy}{dx}+\frac{xy}{x^{2}-1}=\frac{x^{6}+4x}{\sqrt{1-x^{2}}},\ -1<x<1, such that f(0)=0f(0)=0. If 6∫−1/21/2f(x)dx=2π−α6\int_{-1/2}^{1/2} f(x)dx = 2\pi-\alpha, then α2\alpha^{2} is equal to

  7. Question 7

    7 Apr 2025

    138
    Let y=y(x)y=y\left( x \right) be the solution of the differential equation (x2+1)y′−\left( x^{2}+ 1 \right)y^{'}- 2xy=(x4+2x2+1)cos⁡x2xy=\left( x^{4}+ 2x^{2}+ 1 \right)\cos x, y(0)=1y(0) = 1. Then ∫−33y(x)dx\int_{- 3}^{3} y(x)dx is :

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