<p>If <span class="math">\int_0^{100} f(x) dx = a</span>, then <span class="math">\sum_{r=1}^{100} \int_0^1 f(r-1+x) dx</span> equals</p>
<p>(a) <span class="math">100a</span></p>
<p>(b) <span class="math">a</span></p>
<p>(c) <span class="math">0</span></p>
<p>(d) <span class="math">10a</span></p>
Step-by-Step Solution
Key Concept: Use substitution to transform each integral ∫₀¹ f(r-1+x)dx into a form that relates to the original integral ∫₀¹⁰⁰ f(x)dx. The sum of these transformed integrals will cover the entire interval [0,100] exactly once.
Step 1: Consider the general term of the sum:
$$ \int_0^1 f(r-1+x) \, dx $$
where $r$ ranges from $1$ to $100$.
Step 2: Apply the substitution $u = r-1+x$.
Then $du = dx$.
When $x=0$, $u = r-1$.
When $x=1$, $u = r$.
Step 3: Transform the integral using the substitution:
$$ \int_0^1 f(r-1+x) \, dx = \int_{r-1}^r f(u) \, du $$
Step 4: Substitute this transformed integral back into the original sum:
$$ \sum_{r=1}^{100} \int_0^1 f(r-1+x) \, dx = \sum_{r=1}^{100} \int_{r-1}^r f(u) \, du $$
Step 5: Expand the sum and apply the additivity property of definite integrals. The sum represents a concatenation of integrals over consecutive intervals:
$$ \sum_{r=1}^{100} \int_{r-1}^r f(u) \, du = \int_0^1 f(u) \, du + \int_1^2 f(u) \, du + \int_2^3 f(u) \, du + \dots + \int_{99}^{100} f(u) \, du $$
This sum simplifies to a single integral over the entire range:
$$ \int_0^1 f(u) \, du + \int_1^2 f(u) \, du + \dots + \int_{99}^{100} f(u) \, du = \int_0^{100} f(u) \, du $$
Step 6: Given that $\int_0^{100} f(x) \, dx = a$, it follows that:
$$ \sum_{r=1}^{100} \int_0^1 f(r-1+x) \, dx = a $$
Correct Answer: a