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 evaluate both definite integrals and solve the resulting equation for the variable $a$.

Step 2: Detailed Explanation:
Evaluate the first integral:
\[ I_1 = \int_{-5}^5 \frac{1}{20}(x - 5)^2 dx \]
Let \( u = x - 5 \implies du = dx \). Limits: \( x = -5 \to u = -10 \); \( x = 5 \to u = 0 \).
\[ I_1 = \frac{1}{20} \int_{-10}^0 u^2 du = \frac{1}{20} \left[ \frac{u^3}{3} \right]_{-10}^0 = \frac{1}{60} [0 - (-1000)] = \frac{1000}{60} = \frac{50}{3} \]
Evaluate the second integral:
\[ I_2 = \int_5^a \frac{1}{20}(x - 5)^2 dx \]
Let \( v = x - 5 \implies dv = dx \). Limits: \( x = 5 \to v = 0 \); \( x = a \to v = a - 5 \).
\[ I_2 = \frac{1}{20} \int_0^{a-5} v^2 dv = \frac{1}{20} \left[ \frac{v^3}{3} \right]_0^{a-5} = \frac{(a - 5)^3}{60} \]
Given \( I_1 = I_2 \):
\[ \frac{1000}{60} = \frac{(a - 5)^3}{60} \]
\[ (a - 5)^3 = 1000 \]
Taking the cube root of both sides:
\[ a - 5 = 10 \implies a = 15 \]

Step 3: Final Answer:
The value of $a$ is 15.