Question

Assertion (A): The area of the region bounded by the line $y - 1 = x$, the $x$-axis and the ordinates $x = -1$ and $x = 1$ is 2 square units.
Reason (R): The area of the region bounded by the curve $y = f(x)$, the $x$-axis and the ordinates $x = a$ and $x = b$ is given by \[ \int_{a}^{b} f(x)\, dx. \]

• A. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A)
• B. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation for Assertion (A)
• C. Assertion (A) is true, but Reason (R) is false
• D. Assertion (A) is false, but Reason (R) is true

Hint:

Always use definite integrals to calculate the area under a curve between two vertical lines when bounded by the x-axis.

Answer 1

The Correct Option is A

Given: The line is $y = x + 1$.
We need to calculate the area enclosed by this line, the $x$-axis, and the vertical lines $x = -1$ and $x = 1$.
The region is situated between $x = -1$ and $x = 1$, and it is above the $x$-axis.
Therefore, the required area is given by the integral $\int_{-1}^{1} (x + 1) \, dx$.
\[= \left[\dfrac{x^2}{2} + x\right]_{-1}^{1} = \left(\dfrac{1^2}{2} + 1\right) - \left(\dfrac{(-1)^2}{2} + (-1)\right) = \left(\dfrac{1}{2} + 1\right) - \left(\dfrac{1}{2} - 1\right) = \dfrac{3}{2} - (-\dfrac{1}{2}) = 2\]
Thus, the Assertion is correct.
The Reason describes the standard method for calculating the area under a curve using definite integrals.
This method was indeed applied in the Assertion.
Consequently, both the Assertion and the Reason are true, and the Reason accurately explains the Assertion.