Grade 12 · Substitution and bounds · 594 of 600
Substitution and bounds · Exponent factor k=1; Final upper bound B=0.75
Integrate from x=0 to x=0.45. Find the transformed upper bound under u=kx² and the exact integral, then give a decimal approximation. Givens: Exponent factor k=1; Final upper bound B=0.75.
Try the question, use a hint when you need one, and compare your reasoning with the worked solution. The animation starts with this chapter’s values; changing its controls explores a new case.
01 · Make a prediction
Your practice question
Consider the integrand 2(1)x e^((1)x²). The animation's final upper bound is B=0.75. Integrate from x=0 to x=0.45. Find the transformed upper bound under u=kx² and the exact integral, then give a decimal approximation.
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A starting point
With u=kx², du=2kx dx. Transform both endpoints before integrating eᵘ.
Work through the reasoning
Step 1
Identify the model and target
Consider the integrand 2(1)x e^((1)x²). The animation's final upper bound is B=0.75. Integrate from x=0 to x=0.45. The governing relation is ∫₀ᵀ2kx e^(kx²)dx=e^(kT²)−1; u=kx². With u=kx², du=2kx dx. Transform both endpoints before integrating eᵘ.
Step 2
Substitute and calculate
Under u=kx², the upper bound becomes (1)(0.45)²=0.2025. Therefore ∫eᵘdu from 0 to 0.2025 equals e^0.2025−1=0.22446.
Step 3
Check the mathematical meaning
The transformed interval is 0≤u≤0.2025. Its exact integral is e^(0.2025)−1, which is positive because the interval and integrand are positive. Upper x bound: 0.45; Transformed u bound: 0.203; Exact integral: 0.224. Decimal values are rounded, so use unrounded intermediate values.
The answer
Under u=kx², the upper bound becomes (1)(0.45)²=0.2025. Therefore ∫eᵘdu from 0 to 0.2025 equals e^0.2025−1=0.22446. The transformed interval is 0≤u≤0.2025. Its exact integral is e^(0.2025)−1, which is positive because the interval and integrand are positive. Animation check: Upper x bound: 0.45; Transformed u bound: 0.203; Exact integral: 0.224. Decimal displays are rounded; retain the original parameters and exact π until the final step.
Compare the method as well as the result. A different valid method may reach the same answer. Keep exact values until the last step when the question asks for rounding.
02 · Explore the model
See the mathematical relationship move.
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Watch the relationship
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The mathematical idea
Substitution replaces a composite expression and its differential together. Here 2kx dx is exactly du when u=kx². Moving the upper x-bound also moves the upper u-bound, so retaining the original bound after changing variables would integrate over the wrong interval. Consider the integrand 2(1)x e^((1)x²). The animation's final upper bound is B=0.75. Integrate from x=0 to x=0.45. The requested state occurs at 60% playback; use the exact target stated in the question for your calculation.
∫₀ᵀ2kx e^(kx²)dx=e^(kT²)−1; u=kx²
03 · Reflect and transfer
Explain what changes and why.
Why would retaining the old x endpoint after the substitution generally produce an incorrect integral?
This is a distinct guided scenario using a reusable mathematical model. Similar-looking diagrams can represent different given values and conclusions; they are not different mathematical theories.