How to find area between two curves?

Area = ∫[a to b] |f(x) - g(x)| dx where f(x) ≥ g(x). Find intersection points for limits. Integrate difference of upper minus lower function.

What is area ratio divided by vertical line?

Calculate area on left of line and area on right of line separately using definite integrals. Ratio = (Left Area):(Right Area). Simplify to lowest terms.

GCD is greatest common divisor - largest number dividing both m and n. For coprime numbers, gcd = 1.

How to integrate polynomials?

∫x^n dx = x^(n+1)/(n+1) + C. Apply to each term separately.

How to simplify ratio?

Divide both numbers by their GCD to get lowest terms.

Which is upper curve?

Test a point. At x=0: parabola gives y=1, line gives y=3. Line is upper.

How to verify gcd is 1?

Check if numbers share common prime factors. If not, gcd = 1.

Can area be negative?

Take absolute value if negative. Means integration order or limits reversed.

Let the line x = -1 divide the area of the region {(x,y): 1 + x² ≤ y ≤ 3 - x} in the ratio m:n, gcd(m,n) = 1. Then m + n is equal to

Q: How to find intersection points?

Set equations equal: 1 + x² = 3 - x. Rearrange: x² + x - 2 = 0. Factor: (x+2)(x-1) = 0. Solutions: x = -2 and x = 1.

Q: What is definite integral?

∫[a to b] f(x) dx = F(b) - F(a) where F is antiderivative.

Let the line x = -1 divide the area of the region {(x,y): 1 + x² ≤ y ≤ 3 – x} in the ratio m:n, gcd(m,n) = 1. Then m + n is equal to | JEE Main Mathematics

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Math Integration

Q MCQ Area Ratio

Let the line x = -1 divide the area of the region {(x,y): 1 + x² ≤ y ≤ 3 – x} in the ratio m:n, gcd(m,n) = 1. Then m + n is equal to

✅ Correct Answer

✓

Solution Steps

Find intersection points of curves

Parabola: $y = 1 + x^2$

Line: $y = 3 – x$

At intersection: $1 + x^2 = 3 – x$

$x^2 + x – 2 = 0$

$(x + 2)(x – 1) = 0$

$x = -2$ or $x = 1$

Region exists from $x = -2$ to $x = 1$

Identify upper and lower curves

Test at $x = 0$:

Parabola: $y = 1 + 0^2 = 1$

Line: $y = 3 – 0 = 3$

Since $3 > 1$, line is upper curve

Area element: $(3 – x) – (1 + x^2) = 2 – x – x^2$

Calculate left area (from x = -2 to x = -1)

$A_{\text{left}} = \displaystyle\int_{-2}^{-1} (2 – x – x^2) \, dx$

$= \left[2x – \dfrac{x^2}{2} – \dfrac{x^3}{3}\right]_{-2}^{-1}$

At $x = -1$:

$= 2(-1) – \dfrac{(-1)^2}{2} – \dfrac{(-1)^3}{3}$

$= -2 – \dfrac{1}{2} – \dfrac{-1}{3}$

$= -2 – \dfrac{1}{2} + \dfrac{1}{3} = \dfrac{-12 – 3 + 2}{6} = \dfrac{-13}{6}$

At $x = -2$:

$= 2(-2) – \dfrac{(-2)^2}{2} – \dfrac{(-2)^3}{3}$

$= -4 – \dfrac{4}{2} – \dfrac{-8}{3}$

$= -4 – 2 + \dfrac{8}{3} = \dfrac{-12 – 6 + 8}{3} = \dfrac{-10}{3} = \dfrac{-20}{6}$

$A_{\text{left}} = \dfrac{-13}{6} – \dfrac{-20}{6} = \dfrac{7}{6}$

Calculate right area (from x = -1 to x = 1)

$A_{\text{right}} = \displaystyle\int_{-1}^{1} (2 – x – x^2) \, dx$

$= \left[2x – \dfrac{x^2}{2} – \dfrac{x^3}{3}\right]_{-1}^{1}$

At $x = 1$:

$= 2(1) – \dfrac{1^2}{2} – \dfrac{1^3}{3}$

$= 2 – \dfrac{1}{2} – \dfrac{1}{3} = \dfrac{12 – 3 – 2}{6} = \dfrac{7}{6}$

At $x = -1$: (already calculated) $= \dfrac{-13}{6}$

$A_{\text{right}} = \dfrac{7}{6} – \left(\dfrac{-13}{6}\right) = \dfrac{7 + 13}{6} = \dfrac{20}{6} = \dfrac{10}{3}$

Find the ratio m:n

Ratio = $A_{\text{left}} : A_{\text{right}}$

$= \dfrac{7}{6} : \dfrac{10}{3}$

$= \dfrac{7}{6} : \dfrac{20}{6}$

$= 7 : 20$

Check: $\gcd(7, 20) = 1$ ✓ (coprime)

So $m = 7$ and $n = 20$

Calculate m + n

$m + n = 7 + 20 = 27$ ✓

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Final Answer

m + n = 27 (where m:n = 7:20)

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Related Theory

📌 Area Between Two Curves

To find area between curves y = f(x) and y = g(x) from x = a to x = b where f(x) ≥ g(x), use formula: Area = ∫[a to b] (f(x) – g(x)) dx. Steps: First find intersection points by solving f(x) = g(x) to get integration limits. Second determine which function is upper by testing a point in the interval. Third set up integral of (upper – lower). Fourth evaluate using antiderivatives. For vertical line division, calculate left and right areas separately then find their ratio.

📌 Definite Integration Process

Definite integral ∫[a to b] f(x) dx = F(b) – F(a) where F is antiderivative of f. Common antiderivatives: ∫x^n dx = x^(n+1)/(n+1) for n ≠ -1, ∫constant dx = constant × x. Properties include linearity and additivity over intervals. Always evaluate upper limit value minus lower limit value. Be careful with negative signs when substituting negative numbers.

📌 GCD and Coprime Numbers

Greatest Common Divisor (GCD) of two integers is the largest positive integer that divides both. Two numbers are coprime if gcd = 1 meaning they share no common factors except 1. To simplify ratio divide both parts by their GCD. For example 14:21 becomes 2:3 after dividing by gcd(14,21) = 7. When problem states gcd(m,n) = 1 the ratio is already in simplest form.

📌 Parabola and Line Intersection

When parabola y = ax² + bx + c intersects line y = mx + n set equations equal and rearrange to standard quadratic form. Solve for x-coordinates which become integration limits. For our problem: 1 + x² = 3 – x gives x² + x – 2 = 0 which factors as (x+2)(x-1) = 0 yielding x = -2 and x = 1. Always verify solutions by substituting back into both original equations.

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FAQs

How to find area between curves?

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Integrate (upper function – lower function) between intersection points. Find intersections by setting equations equal and solving.

What if integral gives negative?

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Take absolute value. Negative means you subtracted in wrong order (lower – upper) or limits are reversed.

How to divide area by vertical line?

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Calculate left area (from left boundary to dividing line) and right area (from dividing line to right boundary) separately. Find ratio.

What does gcd(m,n) = 1 mean?

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Numbers m and n are coprime – they share no common factors. Ratio is already in simplest form.

How to find intersection points?

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Set equations equal. For 1 + x² = 3 – x: rearrange to x² + x – 2 = 0, factor to get x = -2, 1.

Which curve is upper?

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Test a point. At x = 0: parabola gives y = 1, line gives y = 3. Since 3 > 1, line is upper.

How to evaluate definite integral?

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Find antiderivative F(x). Calculate F(upper limit) – F(lower limit). Be careful with negative numbers.

Can ratio be simplified further?

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Divide both by GCD. For 7:20, check gcd(7,20) = 1 already, so it’s in simplest form.

Why 7:20 not 20:7?

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Convention: left area first. But both give same sum m + n = 27. Problem asks for sum not specific order.

How to verify answer?

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Check: (1) gcd(7,20) = 1 ✓, (2) 7 + 20 = 27 ✓, (3) left area + right area = 7/6 + 10/3 = 27/6 total area ✓

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