If the area of the region $\{(x,y): -1 \le x \le 1, 0 \le y \le f(x)\}$ is $\frac{e^{2}+8e+1}{e}$ , then the value of $a$ is:
- (1) 7
- (2) 6
- (3) 8
- (4) 5
Solution:
$y=f(x)$ is defined as:
$$f(x) = \begin{cases} a & -1 \le x \le 0 \\ a + e^x – e^{-x} & 0 < x \le 1 \end{cases}$$
The area is given by the sum of integrals over these intervals:
$$Area = \int_{-1}^{0} a \, dx + \int_{0}^{1} (a + e^x – e^{-x}) \, dx$$
Calculating the first integral:
$$\int_{-1}^{0} a \, dx = [ax]_{-1}^{0} = a(0) – a(-1) = a$$
Calculating the second integral:
$$\int_{0}^{1} (a + e^x – e^{-x}) \, dx = [ax + e^x + e^{-x}]_{0}^{1}$$
Applying limits:
$$= \left( a(1) + e^1 + e^{-1} \right) – \left( a(0) + e^0 + e^{-0} \right)$$
$$= (a + e + \frac{1}{e}) – (0 + 1 + 1)$$
$$= a + e + \frac{1}{e} – 2$$
Total Area:
$$A_{total} = a + \left( a + e + \frac{1}{e} – 2 \right) = 2a + e + \frac{1}{e} – 2$$
We are given that the area is $\frac{e^2+8e+1}{e}$. We can rewrite this as:
$$\frac{e^2}{e} + \frac{8e}{e} + \frac{1}{e} = e + 8 + \frac{1}{e}$$
Equating the calculated area to the given area:
$$2a + e + \frac{1}{e} – 2 = e + 8 + \frac{1}{e}$$
Subtract $e + \frac{1}{e}$ from both sides:
$$2a – 2 = 8$$
$$2a = 10$$
$$a = 5$$
Ans. (4)