Definite Integrals

In Exercises 5–8, express each limit as a definite integral. Then evaluate the integral to find the value of the limit. In each case, P is a partition of the given interval, and the numbers cₖ are chosen from the subintervals of P.

     n
lim  ∑ (2cₖ - 1)⁻¹/² ∆xₖ, where P is a partition of [1, 5]
∥P∥→0  k = 1

검증된 단계별 안내

Recognize that the given limit represents a Riemann sum of the form \(\lim_{\|P\| \to 0} \sum_{k=1}^n f(c_k) \Delta x_k\), which corresponds to the definite integral of \(f(x)\) over the interval \([a, b]\).

Identify the function inside the sum: \(f(c_k) = (2c_k - 1)^{-1/2}\), and the interval of integration is \([1, 5]\).

Express the limit as the definite integral: \(\int_1^5 (2x - 1)^{-1/2} \, dx\).

To evaluate the integral, use a substitution method. Let \(u = 2x - 1\), then compute \(du = 2 \, dx\), which implies \(dx = \frac{du}{2}\).

Change the limits of integration accordingly: when \(x=1\), \(u=2(1)-1=1\); when \(x=5\), \(u=2(5)-1=9\). Rewrite the integral in terms of \(u\) and integrate \(\int_1^9 u^{-1/2} \cdot \frac{1}{2} \, du\).

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주요 개념

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Definite Integral as a Limit of Riemann Sums

A definite integral can be defined as the limit of Riemann sums, where the sum of function values at chosen points multiplied by subinterval widths approaches the exact area under the curve as the partition gets finer. This concept connects sums to integrals and is fundamental for interpreting limits of sums as integrals.

Partition and Sample Points in Integration

A partition divides the interval into subintervals, and sample points (cₖ) are chosen within these subintervals to evaluate the function. The choice of sample points affects the Riemann sum, but as the norm of the partition approaches zero, the sum converges to the definite integral regardless of these choices.

Evaluating Definite Integrals

Once the limit is expressed as a definite integral, evaluating it involves finding an antiderivative of the integrand and applying the Fundamental Theorem of Calculus. This process yields the exact value of the integral, which corresponds to the original limit of the sum.