Question

If $f(a)$ is the area bounded in the first quadrant by $x=0$, $x=1$, $y=x^2$ and $y=|ax-5|-|1-ax|+ax^2$, then find $f(0)+f(1)$.

• A. $\dfrac{11}{3}$
• B. $\dfrac{13}{3}$
• C. $\dfrac{17}{3}$
• D. $\dfrac{23}{3}$

Hint:

Always simplify modulus expressions interval-wise before integrating.

Answer 1

The Correct Option is D

To solve the problem, we need to find the area \(f(a)\) bounded in the first quadrant by the curves \(x=0\), \(x=1\), \(y=x^2\), and \(y=|ax-5|-|1-ax|+ax^2\), and then compute \(f(0) + f(1)\).

1. First, consider \(f(0)\): 

For \(a=0\), the function becomes:

• \(y = |-5| - |1| + 0 = 5 - 1 = 4\)

We now find the area between \(y = x^2\) and \(y = 4\) from \(x = 0\) to \(x = 1\):

The area \(A_0\) is given by:

• \(A_0 = \int_{0}^{1} (4 - x^2) \, dx\)
• Integrating, we have:
• \(\left[ 4x - \frac{x^3}{3} \right]_{0}^{1} = \left( 4 \cdot 1 - \frac{1^3}{3} \right) - (0) = 4 - \frac{1}{3} = \frac{12}{3} - \frac{1}{3} = \frac{11}{3}\)

1. Now consider \(f(1)\):

For \(a=1\), the function becomes:

• \(y = |x-5| - |1-x| + x^2\)

Examining the function:

• For \(0 \leq x < 1\), we have \(|x-5| = 5-x\) and \(|1-x| = 1-x\).
• The function simplifies to:
• \(y = (5-x) - (1-x) + x^2 = 4 + x^2\)

Thus, for this range, the area \(A_1\) is given by:

• \(A_1 = \int_{0}^{1} ((4+x^2) - x^2) \, dx = \int_{0}^{1} 4 \, dx = [4x]_{0}^{1} = 4(1) - 4(0) = 4\)

Hence, \(f(1)=4\).

1. Finally, calculate \(f(0) + f(1)\):

• \(f(0) + f(1) = \frac{11}{3} + 4 = \frac{11}{3} + \frac{12}{3} = \frac{23}{3}\)

Thus, the value of \(f(0) + f(1)\) is the correct option:

\(\frac{23}{3}\)