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 $f(0)+f(1)$.

Hint:

Always simplify modulus expressions interval-wise before integrating.

Answer 1

Correct Answer: 8

Step 1: Evaluate $f(0)$.
For $a=0$: \[ y=| -5|-|1|+0=5-1=4 \] Area between $y=x^2$ and $y=4$ from $0$ to $1$: \[ f(0)=\int_0^1(4-x^2)\,dx \] \[ =\left[4x-\frac{x^3}{3}\right]_0^1=\frac{11}{3} \] Step 2: Evaluate $f(1)$.
For $a=1$: \[ y=|x-5|-|1-x|+x^2 \] In first quadrant $0\le x\le1$: \[ |x-5|=5-x,\quad |1-x|=1-x \] \[ y=5-x-(1-x)+x^2=4+x^2 \] Area between $y=x^2$ and $y=4+x^2$: \[ f(1)=\int_0^1 4\,dx=4 \] Step 3: Final calculation.
\[ f(0)+f(1)=\frac{11}{3}+4=\frac{23}{3} \] Final conclusion.
The required value is $\dfrac{23}{3}$.