The area bound by the graph of f(x)=ex−1 f(x)=e^x-1 , the x-axis and the line x=2 is equal to:

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We start by identifying the region whose area is asked for: it lies under the curve f(x) = e^x - 1, above the x-axis, and limited on the right by the vertical line x = 2. First, find where the curve intersects the x-axis to set the lower limit of integration: e^x - 1 = 0 implies e^x = 1, so x = 0. Therefore, the area is the integral from 0 to 2 of (e^x - 1) dx.

Option a: e^2 - 2. This would correspond to integrating e^x alone from 0 to 2, giving e^2 - 1, and subtracting 1 gives e^2 - 2 only if you misaccounted the constant, but the correct evaluation is e^2 - 3, so this is incorrect.

Option b: e^2 + 1. This would arise from an incorrect sign or incorrect lower limit, since the actual integral yields e^2 - 3, not e^2 + 1. The presence of +1 instead of -3 indicates a miscalculation of the constant term.

Option c: e^2 + 2. Similar to option b, this option incorrectly adds a positive constant that does not come from the definite integral ∫0^2 (e^x - 1) dx, where the constant term should be negative due to the -1 in the integrand.

Option d: e^2 - 3. This matches the correct evaluation. Compute step by step: ∫0^2 (e^x - 1) dx = [e^x]0^2 - [x]0^2 = (e^2 - 1) - (2 - 0) = e^2 - 1 - 2 = e^2 - 3. The result e^2 - 3 is the exact area under the curve from x = 0 to x = 2, above the x-axis, bounded by x = 2.

In summary, the correct area is obtained by integrating e^x - 1 from 0 to 2, yielding e^2 - 3, so option d is the correct choice. The other options reflect common mistakes such as omitting the lower limit contribution, mismanaging the constant term from the -1, or miscomputing the definite integral.

MUF0092 Mathematics Unit 2 - Semester 2, 2025 3.1 Integration checking understanding

The area bound by the graph of f(x)=ex−1 f(x)=e^x-1 , the x-axis and the line x=2 is equal to:

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