Question:
The area (in sq. units) of the region bounded between the curve \( y=x^{2}+5x+1 \) and the line \( 7x-y+1=0 \) is
The area (in sq. units) of the region bounded between the curve \( y=x^{2}+5x+1 \) and the line \( 7x-y+1=0 \) is
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Archimedes' Formula Shortcut: The area enclosed by a parabola and a line cutting it at two points is always equal to \( \frac{|a|}{6}(\beta - \alpha)^3 \). Here, \( \frac{1}{6}(2 - 0)^3 = \frac{8}{6} = \frac{4}{3} \). You can find the answer in under 5 seconds!
Updated On: Jun 8, 2026
- \( 2 \)
- \( \frac{3}{4} \)
- \( \frac{4}{3} \)
- \( \frac{2}{5} \)
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The Correct Option is C
Solution and Explanation
Concept:
The area bounded between a line and a quadratic curve is given by \( \int_{\alpha}^{\beta} (y_{\text{upper}} - y_{\text{lower}}) \, dx \), where \( \alpha, \beta \) are the \( x \)-coordinates of their intersection points.
Step 1: Finding the intersection points.
From the line equation, \( y = 7x + 1 \). Equating this with the curve equation: \[ 7x + 1 = x^2 + 5x + 1 \implies x^2 - 2x = 0 \implies x(x-2) = 0 \] Thus, the intersection boundary points are \( x = 0 \) and \( x = 2 \).
Step 2: Setting up the area integral expression.
In the interval \( (0, 2) \), testing \( x = 1 \) shows that the line (\( y=8 \)) lies above the curve (\( y=7 \)). \[ \text{Area} = \int_{0}^{2} [(7x+1) - (x^2+5x+1)] \, dx = \int_{0}^{2} (2x - x^2) \, dx \]
Step 3: Evaluating the integral.
\[ \text{Area} = \left[ x^2 - \frac{x^3}{3} \right]_{0}^{2} = \left( 2^2 - \frac{2^3}{3} \right) - 0 = 4 - \frac{8}{3} = \frac{12 - 8}{3} = \frac{4}{3} \] This matches option (C) perfectly.
Step 1: Finding the intersection points.
From the line equation, \( y = 7x + 1 \). Equating this with the curve equation: \[ 7x + 1 = x^2 + 5x + 1 \implies x^2 - 2x = 0 \implies x(x-2) = 0 \] Thus, the intersection boundary points are \( x = 0 \) and \( x = 2 \).
Step 2: Setting up the area integral expression.
In the interval \( (0, 2) \), testing \( x = 1 \) shows that the line (\( y=8 \)) lies above the curve (\( y=7 \)). \[ \text{Area} = \int_{0}^{2} [(7x+1) - (x^2+5x+1)] \, dx = \int_{0}^{2} (2x - x^2) \, dx \]
Step 3: Evaluating the integral.
\[ \text{Area} = \left[ x^2 - \frac{x^3}{3} \right]_{0}^{2} = \left( 2^2 - \frac{2^3}{3} \right) - 0 = 4 - \frac{8}{3} = \frac{12 - 8}{3} = \frac{4}{3} \] This matches option (C) perfectly.
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