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: Compute f(0)

Substitute a = 0 in the given expression:

y = |−5| − |1| + 0

y = 5 − 1 = 4

Thus, the curves considered are y = 4 and y = x² on the interval [0, 1].

Area enclosed between them:

f(0) = ∫₀¹ (4 − x²) dx

f(0) = [4x − x³/3]₀¹

f(0) = 11/3

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Step 2: Compute f(1)

Put a = 1 in the given expression:

y = |x − 5| − |1 − x| + x²

For 0 ≤ x ≤ 1:

|x − 5| = 5 − x,   |1 − x| = 1 − x

Hence,

y = (5 − x) − (1 − x) + x²

y = 4 + x²

Now the curves are y = x² and y = 4 + x².

Vertical separation between them is constant and equal to 4.

So,

f(1) = ∫₀¹ 4 dx

f(1) = 4

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Step 3: Required sum

f(0) + f(1) = 11/3 + 4

= 23/3

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Final Answer:

The required value is
23/3