Integral Insight Solution Guide
Integral Insight
Function Composition & The Fundamental Theorem
AP Calculus AB
The Challenge
Consider the function \( f \) defined by the integral:
\[ f(x) = \int_{2}^{x} (3t + 2) \, dt \]
Find the value of the composition: \( f(3 \cdot f(3)) \)
Phase 1
Evaluate \( f(3) \)
Phase 2
Scale Argument
Phase 3
Final Result
Step 1: Evaluation of \( f(3) \)
Inner Component
Apply the Fundamental Theorem of Calculus with the upper bound \( x = 3 \):
\[ f(3) = \int_{2}^{3} (3t + 2) \, dt = \left[ 1.5t^2 + 2t \right]_{2}^{3} \]
At \( t = 3 \)
\( 1.5(3)^2 + 2(3) = 13.5 + 6 = \mathbf{19.5} \)
At \( t = 2 \)
\( 1.5(2)^2 + 2(2) = 6 + 4 = \mathbf{10} \)
\( f(3) = 19.5 - 10 = 9.5 \)
Step 2: Input Scaling
Find the value to be used as the new upper limit:
\( 3 \times 9.5 = 28.5 \)
Step 3: Final Integration \( f(28.5) \)
Outer Component
Integrate from the original lower bound (2) to the new upper bound (28.5):
\[ f(28.5) = \int_{2}^{28.5} (3t + 2) \, dt = \left[ 1.5t^2 + 2t \right]_{2}^{28.5} \]
Arithmetic Breakdown
Value at 28.5: \( 1.5(28.5)^2 + 2(28.5) = 1275.375 \)
Value at 2.0: \( 1.5(2)^2 + 2(2) = 10.000 \)
Final Difference: 1265.375
The Solution
1265.375
\( f(3 \cdot f(3)) = 1265 \frac{3}{8} \)
Verification: The Trapezoid Shortcut
Since the integrand is linear, the integral represents the area of a trapezoid. This is a great way to verify your integration!
Dimensions
Height (h) = 26.5
Base1 (at 2) = 8
Base2 (at 28.5) = 87.5
Area = ½h(b1 + b2)
\( 0.5 \times 26.5 \times (8 + 87.5) = \mathbf{1265.375} \)
