Approximating Integrals: Fresh Take

Understand that while all differentiable functions can be derived in a straightforward formula, not all functions can be integrated into a simple antiderivative
Calculate bounds for the area calculations under curves when direct integration methods aren’t applicable
Explain how estimating the bounds of an integral affects the accuracy of the approximation

Approximating Integrals

The Main Idea

Not all functions have antiderivatives in terms of elementary functions
Riemann sums can be used to approximate definite integrals
Upper and lower bounds provide an interval for the true value
Error estimation helps quantify the accuracy of approximations
Non-integrable Functions:
- Some functions don’t have antiderivatives in terms of elementary functions
- Example:
  - $g(x) = \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}$
Riemann Sum Approximation:
- Left-endpoint sum:
  - $L_n = \sum_{i=1}^n f(\frac{i-1}{n}) \cdot \frac{1}{n}$
- Right-endpoint sum:
  - $R_n = \sum_{i=1}^n f(\frac{i}{n}) \cdot \frac{1}{n}$
Bounds:
- For decreasing functions:
  - $R_n \leq \text{True Value} \leq L_n$
- For increasing functions:
  - $L_n \leq \text{True Value} \leq R_n$
Error Estimation:
- Maximum error:
  - $\varepsilon = \frac{\text{Upper Bound} - \text{Lower Bound}}{2}$

Consider the definite integral:

$\int_{-1}^0 \sqrt{1-x^4} dx$

Can this integral be evaluated using the Fundamental Theorem of Calculus, Part 2?
If not, provide an upper bound and a lower bound for the true value of the definite integral using $10$ subintervals.
Choose a single value from that interval to use as an approximation. What is the maximum error that could be associated with that approximation?

No, this integral cannot be evaluated using the Fundamental Theorem of Calculus, Part 2. The function $f(x) = \sqrt{1 - x^4}$ does not have an elementary antiderivative.
We’ll use Riemann sums with $n = 10$ subintervals to find upper and lower bounds.
Left-endpoint sum (upper bound):

$L_{10} = \sum_{i=1}^{10} \sqrt{1 - (-1 + (i-1) \cdot 0.1)^4} \cdot 0.1 \approx 0.360$

Right-endpoint sum (lower bound):

$R_{10} = \sum_{i=1}^{10} \sqrt{1 - (-1 + i \cdot 0.1)^4} \cdot 0.1 \approx 0.332$

Therefore, $0.332 \leq \int_{-1}^0 \sqrt{1-x^4} dx \leq 0.360$
Let’s choose the midpoint of the interval $[0.332, 0.360]$ as our approximation:
$\text{Approximation} = \frac{0.332 + 0.360}{2} = 0.346$

The maximum error is the distance from the midpoint to either endpoint:

$\text{Maximum Error} = 0.360 - 0.346 = 0.346 - 0.332 = 0.014$

Thus, our approximation is $0.346$ , and it could be off by at most $0.014$ from the true value of the integral.