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18. Using integration, find the area of the triangular region whose sides have the equations y = 2x +1, y = 3x + 1 and x = 4.

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    Answered by

    Vishal Baghel | Contributor-Level 10

    4 months ago

    The given equation of the sides of triangle is

    y=2x+1 --------------------(1)

    y=3x+1 -------------------(2)

    x=4 -------------------------(3)

    Solving eqn (1) and (2) for x & y we get

    3x+1=2x+1=3x−2x=1−1=x=0&y=2×0+1=1

    ∴ The point of inersection of line (1)and (2)is A (0,1)

    Putting x=4 in eq (1) and (2)we get,

    y=2×4+1=8+1=9&y=3×4+1=12+1=13

    ∴ The point of intersection of line (1)and (3) is B(4,9) and C (4,13)

    Hence the required area enclosed ABC

    = ∫ 0 4 y l i n e ( 2 ) d x − ∫ 0 4 y l i n e ( 1 ) d x = ∫ 0 4 [ 3 x + 1 ] d x − ∫ 0 4 [ 2 x + 1 ] d x = [ 3 x 2 2 + x ] 0 4 − [ 2 x 2 2 + x ] 0 4 = [ ( 3 2 ( 4 ) 2 + 4 ) − ( 3 × 0 2 2 + 0 ) ] − [ ( 4 2 + 4 ) − ( 0 2 + ) ] = 2 4 + 4 − 2 0 = 8 u n i t 2

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Let a cure y = y(x) pass through the point (3, 3) and the area of the origin under this curve, above the x-axis and between the abscissae 3 and x (>3) be  ( y x ) 3 . If the curve also passes through the point ( α ,     6 1 0 )  in the first quadrant, then a is equal to………….
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Raj Pandey

∫ 3 x f ( x ) d x = ( f ( x ) x ) 3 ⇒ x 3 ∫ 3 x f ( x ) d x = f 3 ( x ) , differentiating w.r.to x

x 3 f ( x ) + 3 x 2 f 3 ( x ) x 3 = 3 f 2 ( x ) f ' ( x ) ⇒ 3 y 2 d y d x = x 3 y = 3 y 3 x ⇒ 3 x y d y d x = x 4 + 3 y 2  

After solving we get  y 2 = x 4 3 + c x 2  also curve passes through (3, 3) Þ c = -2


∴ y 2 = x 4 3 − 2 x 2 which passes through ( α , 6 1 0 ) ∴ α 4 − 6 α 2 3 = 3 6 0 ⇒ α = 6  

The odd natural number a, such that the area of the region bounded by y = 1, y = 3, x = 0, x = ya is 3 6 4 3 ,  is equal to:
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Raj Pandey

Since a is a odd natural number then | ∫ 1 3 y a d y | = 3 6 4 3 ⇒ | ( y a + 1 a + 1 ) 1 3 | = 3 6 4 3 ⇒ 3 a + 1 a + 1 = 3 6 4 3  

 Þ a = 5

The area bounded by the curves y2 = x and x2 = -y is equal to
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y 2 = x , a = 1 4

x 2 = − y , b = 1 4

A = 1 6 | a b | 3 = 1 3

limₓ→∞ [∫₀ˣ tan⁻¹t dt] / √x²+1 is equal to:
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lim (x→∞) (∫? ^ (√x²+1) tan? ¹t dt) / x = lim (x→∞) (tan? ¹ (√x²+1) * (x/√ (x²+1) = lim (x→∞) (tan? ¹ x) * (x/√ (x²+1) = π/2

Area of the region bounded by x = 0, y = 0, x = 2, y = 2, y ≤ e? and y ≥ ln x, is:
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A = ∫? ² lnx dx = 2ln2 – 1
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