Definite Integration
Properties of Definite Integrals
Grade 12
Question:
<p><span class='math'>\sum_{r=1}^{10} \int_{0}^{1} f(r-1+x)\,dx</span> is equal to</p>
<p>(a) <span class='math'>\int_{0}^{1} f(x)\,dx</span></p>
<p>(b) <span class='math'>\int_{0}^{1} f(x)\,dx</span></p>
<p>(c) <span class='math'>10\int_{0}^{1} f(x)\,dx</span></p>
<p>(d) <span class='math'>9\int_{0}^{1} f(x)\,dx</span></p>
Step-by-Step Solution
Key Concept: Use substitution in summation of integrals; recognize telescoping or additive property
<p>By substitution <span class='math'>u = r-1+x</span>, each integral <span class='math'>\int_{0}^{1} f(r-1+x)\,dx = \int_{r-1}^{r} f(u)\,du</span>. Summing from r=1 to 10 gives <span class='math'>\int_{0}^{10} f(u)\,du</span> if f is continuous. For the form shown, assuming additivity: <span class='math'>10\int_{0}^{1} f(x)\,dx</span>.</p>
Correct Answer: C