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Mathematics - Area Under Curves Question with Solution | TestHub

MathematicsArea Under CurvesArea of Curve along axisEasy2 minQB

MathematicsEasysingle choice

Passage / Comprehension

A curve passing through the origin is such that the slope of the tangent at any point is the reciprocal of the sum of the co-ordinates of the point of tangency.

The area bounded by the curve and the abscissae $y = 0 and y = 1 (and the y-axis) is$

Options:

Answer:

Solution:

$\frac{d y}{d x} = \frac{1}{x + y}$ , so $\frac{d x}{d y} - x = y$ . Solving with $x (0) = 0 : x = e^{y} - y - 1$ (equivalently $y = ln (x + y + 1)$ ). Area $= \int_{0}^{1} (e^{y} - y - 1) d y = (e - 1) - \frac{1}{2} - 1 = e - \frac{5}{2}$ . (Source key was B; the correct value is $e - \frac{5}{2}$ , option C.)

Stream:JEESubject:MathematicsTopic:Area Under CurvesSubtopic:Area of Curve along axis

⏱ 2mℹ️ Source: QB

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