Q. Let \(f(\alpha)\) denote the area of the region in the first quadrant bounded by
\(x = 0,\ x = 1,\ y^2 = x\) and
\(y = |\alpha x - 5| - |1 - \alpha x| + \alpha x^2\).
Then \((f(0) + f(1))\) is equal to
Explanation
First evaluate \(f(0)\).
For \(\alpha = 0\),
\[
y = | -5 | - |1| + 0 = 5 - 1 = 4
\]
So the region is bounded by \(y = 4\) and \(y = \sqrt{x}\) from \(x = 0\) to \(x = 1\).
\[
f(0) = \int_0^1 (4 - \sqrt{x})\,dx
\]
\[
= \left[4x - \frac{2}{3}x^{3/2}\right]_0^1
= 4 - \frac{2}{3}
= \frac{10}{3}
\]
Now evaluate \(f(1)\).
For \(\alpha = 1\),
\[
y = |x - 5| - |1 - x| + x^2
\]
For \(0 \le x \le 1\),
\[
|x-5| = 5 - x,\quad |1-x| = 1 - x
\]
\[
y = (5-x) - (1-x) + x^2 = 4 + x^2
\]
Thus,
\[
f(1) = \int_0^1 (4 + x^2 - \sqrt{x})\,dx
\]
\[
= \left[4x + \frac{x^3}{3} - \frac{2}{3}x^{3/2}\right]_0^1
\]
\[
= 4 + \frac{1}{3} - \frac{2}{3} = \frac{11}{3}
\]
Hence,
\[
f(0) + f(1) = \frac{10}{3} + \frac{11}{3} = 7
\]
Therefore, the correct answer is 7.