Question

Let $f(x) = \frac{1}{20}(x - 5)^2, x \in \mathbb{R}$. If $\int_{-5}^{5} f(x) dx = \int_5^{a} f(x) dx$, where $a > 5$ is a real constant, then the value of $a$ is equal to

• A. 10
• B. 12
• C. 15
• D. 18
• E. 20

Hint:

Substitution before integration is often cleaner. Note that the vertex of the parabola is at \( x=5 \). The integral from \( -5 \) to \( 5 \) covers a distance of 10 from the vertex. For the other side to be equal, it must also cover a distance of 10 from the vertex starting at \( x=5 \). Thus, \( a = 5 + 10 = 15 \).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We must equate two definite integrals of the same polynomial function. By evaluating both sides, we form an equation to solve for the upper bound \(a\).
Step 2: Key Formula or Approach:
Integrate \(f(x)\): \(\int (x - 5)^2 dx = \frac{(x - 5)^3}{3}\).
Evaluate the definite integrals for both intervals.
Step 3: Detailed Explanation:
Evaluate the left-hand side (LHS):
\[ \text{LHS} = \frac{1}{20} \int_{-5}^{5} (x - 5)^2 dx \] \[ = \frac{1}{20} \left[ \frac{(x - 5)^3}{3} \right]_{-5}^{5} = \frac{1}{60} \left[ (5 - 5)^3 - (-5 - 5)^3 \right] \] \[ = \frac{1}{60} [0 - (-10)^3] = \frac{1}{60} (1000) = \frac{100}{6} = \frac{50}{3} \] Evaluate the right-hand side (RHS):
\[ \text{RHS} = \frac{1}{20} \int_{5}^{a} (x - 5)^2 dx \] \[ = \frac{1}{20} \left[ \frac{(x - 5)^3}{3} \right]_{5}^{a} = \frac{1}{60} \left[ (a - 5)^3 - (5 - 5)^3 \right] \] \[ = \frac{1}{60} (a - 5)^3 \] Equate LHS and RHS:
\[ \frac{1000}{60} = \frac{(a - 5)^3}{60} \] Multiply both sides by 60:
\[ 1000 = (a - 5)^3 \] Take the cube root of both sides:
\[ \sqrt[3]{1000} = a - 5 \] \[ 10 = a - 5 \implies a = 15 \] Step 4: Final Answer:
The value of \(a\) is 15.