Question:
Which ONE of the following options is the closest approximation to the value of the given definite integral?
\[ \int_0^1 e^{-x^2} dx \]
Note: Use numerical integration
Which ONE of the following options is the closest approximation to the value of the given definite integral?
\[ \int_0^1 e^{-x^2} dx \]
Note: Use numerical integration
\[ \int_0^1 e^{-x^2} dx \]
Note: Use numerical integration
Show Hint
e^-x^2 has no simple antiderivative. Use the trapezoidal rule with f(0)=1 and f(1)=1/e over the single interval [0,1].
Updated On: Aug 5, 2026
- \( 1 \)
- \( \frac{1}{e} \)
- \( \frac{e+1}{2e} \)
- \( \frac{1+e}{2} \)
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Question:
The function \( e^{-x^2} \) has no simple antiderivative in terms of standard functions, so we cannot integrate it directly using normal rules.
The question tells us to use numerical integration instead, to get an approximate value over the interval from 0 to 1.
Step 2: Pick a numerical method:
The simplest numerical method for a single interval is the trapezoidal rule, which approximates the area under a curve as the area of a trapezoid.
\[ I \approx \frac{h}{2} \left[ f(a) + f(b) \right] \]
Here \( h \) is the width of the interval, and \( f(a) \), \( f(b) \) are the function values at the two end points.
Step 3: Compute the function values at the ends:
Since \( a = 0 \) and \( b = 1 \), the width is \( h = 1 \).
\[ f(0) = e^{-0^2} = e^0 = 1, \quad f(1) = e^{-1^2} = e^{-1} = \frac{1}{e} \]
Step 4: Apply the trapezoidal formula:
\[ I \approx \frac{1}{2} \left[ 1 + \frac{1}{e} \right] = \frac{1}{2} \times \frac{e+1}{e} = \frac{e+1}{2e} \]
Step 5: Check option (A) 1.
This is just the value of \( f(0) \) alone, not an average over the interval, so it does not represent the approximate integral.
Step 6: Check option (B) \( \frac{1}{e} \).
This is just \( f(1) \) alone, again only one end point value and not the trapezoidal average, so this option is wrong.
Step 7: Check option (C) \( \frac{e+1}{2e} \).
This matches our trapezoidal rule result exactly, so this is the correct approximation.
Step 8: Check option (D) \( \frac{1+e}{2} \).
This looks close but it is missing the division by \( e \) inside the average, since it directly averages 1 and \( e \) instead of 1 and \( \frac{1}{e} \), so this option is wrong.
Final Answer:
The trapezoidal rule gives the closest approximation as \( \frac{e+1}{2e} \). \[ \boxed{\frac{e+1}{2e}} \]
The function \( e^{-x^2} \) has no simple antiderivative in terms of standard functions, so we cannot integrate it directly using normal rules.
The question tells us to use numerical integration instead, to get an approximate value over the interval from 0 to 1.
Step 2: Pick a numerical method:
The simplest numerical method for a single interval is the trapezoidal rule, which approximates the area under a curve as the area of a trapezoid.
\[ I \approx \frac{h}{2} \left[ f(a) + f(b) \right] \]
Here \( h \) is the width of the interval, and \( f(a) \), \( f(b) \) are the function values at the two end points.
Step 3: Compute the function values at the ends:
Since \( a = 0 \) and \( b = 1 \), the width is \( h = 1 \).
\[ f(0) = e^{-0^2} = e^0 = 1, \quad f(1) = e^{-1^2} = e^{-1} = \frac{1}{e} \]
Step 4: Apply the trapezoidal formula:
\[ I \approx \frac{1}{2} \left[ 1 + \frac{1}{e} \right] = \frac{1}{2} \times \frac{e+1}{e} = \frac{e+1}{2e} \]
Step 5: Check option (A) 1.
This is just the value of \( f(0) \) alone, not an average over the interval, so it does not represent the approximate integral.
Step 6: Check option (B) \( \frac{1}{e} \).
This is just \( f(1) \) alone, again only one end point value and not the trapezoidal average, so this option is wrong.
Step 7: Check option (C) \( \frac{e+1}{2e} \).
This matches our trapezoidal rule result exactly, so this is the correct approximation.
Step 8: Check option (D) \( \frac{1+e}{2} \).
This looks close but it is missing the division by \( e \) inside the average, since it directly averages 1 and \( e \) instead of 1 and \( \frac{1}{e} \), so this option is wrong.
Final Answer:
The trapezoidal rule gives the closest approximation as \( \frac{e+1}{2e} \). \[ \boxed{\frac{e+1}{2e}} \]
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