Key Concepts and Formulas

• Definite Integral as the Limit of a Sum (Riemann Sum): The limit of a sum can be expressed as a definite integral using the formula: $\mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 1}^n f\left( {{r \over n}} \right) = \int\limits_0^1 f(x) dx$ This formula is particularly useful when dealing with limits involving sums of terms that depend on $n$ and an index $r$.

• Power Rule for Integration: The integral of $x^p$ with respect to $x$ is given by: $\int {{x^p}} dx = {{{x^{p + 1}}} \over {p + 1}} + C \quad (\text{for } p \neq -1)$

• Evaluation of Definite Integrals: The definite integral of a function $f(x)$ from $a$ to $b$ is evaluated as $F(b) - F(a)$, where $F(x)$ is the antiderivative of $f(x)$. $\int\limits_a^b f(x) dx = [F(x)]_a^b = F(b) - F(a)$

Step-by-Step Solution

Step 1: Rewrite the Given Expression in the Form of a Riemann Sum

We are asked to find the limit: $L = \mathop {\lim }\limits_{n \to \infty } {{{1^p} + {2^p} + {3^p} + ..... + {n^p}} \over {{n^{p + 1}}}}$ First, we express the sum in sigma notation: $L = \mathop {\lim }\limits_{n \to \infty } {{\sum\limits_{r = 1}^n {{r^p}} } \over {{n^{p + 1}}}}$ To match the Riemann sum formula, we need to isolate a factor of $1/n$ and have the remaining terms as a function of $r/n$. We can rewrite the denominator as $n^p \cdot n$: $L = \mathop {\lim }\limits_{n \to \infty } {{\sum\limits_{r = 1}^n {{r^p}} } \over {{n^p} \cdot n}}$ Now, we can factor out $1/n$ and distribute $n^p$ to each term inside the summation by dividing $r^p$ by $n^p$: $L = \mathop {\lim }\limits_{n \to \infty } {1 \over n} \sum\limits_{r = 1}^n {{{r^p}} \over {{n^p}}}$ Using the property of exponents $(a/b)^m = a^m / b^m$, we can rewrite the term inside the summation as $(r/n)^p$: $L = \mathop {\lim }\limits_{n \to \infty } {1 \over n} \sum\limits_{r = 1}^n {\left( {{r \over n}} \right)^p}$ This expression is now in the standard form of a Riemann sum: $\mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 1}^n f\left( {{r \over n}} \right)$.

Step 2: Convert the Riemann Sum to a Definite Integral

By comparing our expression with the Riemann sum formula, we can identify the function $f(x)$ and the limits of integration. We have $f\left( {{r \over n}} \right) = \left( {{r \over n}} \right)^p$. By setting $x = r/n$, we find that the function is $f(x) = x^p$.

The term $1/n$ outside the summation corresponds to $dx$ in the integral. The limits of integration are determined by the range of $r/n$ as $n \to \infty$:

• The lower limit: As $r=1$ and $n \to \infty$, $r/n = 1/n \to 0$.
• The upper limit: As $r=n$ and $n \to \infty$, $r/n = n/n = 1$.

Therefore, the limit of the sum can be converted into the definite integral: $L = \int\limits_0^1 {{x^p}} dx$

Step 3: Evaluate the Definite Integral

We now evaluate the definite integral $\int\limits_0^1 {{x^p}} dx$ using the power rule for integration. $\int\limits_0^1 {{x^p}} dx = \left[ {{{x^{p + 1}}} \over {p + 1}} \right]_0^1$ To evaluate, we substitute the upper limit ($x=1$) and subtract the result of substituting the lower limit ($x=0$): $L = \left( {{{1^{p + 1}}} \over {p + 1}} \right) - \left( {{{0^{p + 1}}} \over {p + 1}} \right)$ Assuming $p+1 \neq 0$ (which is consistent with the options provided), and knowing that $1^{p+1} = 1$ and $0^{p+1} = 0$ (for $p+1 > 0$), we get: $L = {1 \over {p + 1}} - 0$ $L = {1 \over {p + 1}}$

Step 4: Match with Options

The calculated value of the limit is $1/(p+1)$. Comparing this result with the given options, we find that it matches option (A).

Common Mistakes & Tips

• Incorrectly identifying $f(x)$: Ensure that the expression inside the summation can be written as $f(r/n)$. Sometimes, algebraic manipulation is needed to achieve this form.
• Mistakes in integration: Double-check the application of the power rule for integration, especially for the exponent $p$. Remember that the formula changes if $p=-1$.
• Determining integration limits: The limits of integration are crucial and are derived from the behavior of $r/n$ as $n \to \infty$ for the smallest and largest values of $r$ in the sum.

Summary

The problem requires evaluating a limit of a sum, which can be effectively solved by recognizing it as a Riemann sum and converting it into a definite integral. By rewriting the given expression into the form $\mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 1}^n f\left( {{r \over n}} \right)$, we identified $f(x) = x^p$. This sum was then transformed into the definite integral $\int_0^1 x^p dx$. Evaluating this integral using the power rule yielded the result $1/(p+1)$.

The final answer is \boxed{\frac{1}{p+1}}. This corresponds to option (A).