Calculate integral from 0 to 10 a^x dx by forming

Calculate $\int_0^{10} a^x \, dx$ by forming the Riemann sum.

We will evaluate the integral using the definition as the limit of Riemann sums.

Step 1: Partition the interval

Divide the interval $[0,10]$ into $n$ equal subintervals. The width of each subinterval is:

\[\Delta x = \frac{10-0}{n} = \frac{10}{n}\]

The partition points are:

\[x_i = 0 + i \cdot \Delta x = \frac{10i}{n}, \quad i = 0, 1, 2, \ldots, n\]

Step 2: Choose the sample points

We use the right endpoint of each subinterval:

\[\xi_i = x_i = \frac{10i}{n}\]

Step 3: Form the Riemann sum

The Riemann sum is:

\[S_n = \sum_{i=1}^{n} f(\xi_i) \Delta x = \sum_{i=1}^{n} a^{10i/n} \cdot \frac{10}{n} = \frac{10}{n} \sum_{i=1}^{n} a^{10i/n}\]

Step 4: Simplify the sum

\[S_n = \frac{10}{n} \sum_{i=1}^{n} a^{10i/n} = \frac{10}{n} \sum_{i=1}^{n} \left(a^{10/n}\right)^i\]

This is a geometric series with first term $r = a^{10/n}$ and common ratio $r = a^{10/n}$:

\[\sum_{i=1}^{n} \left(a^{10/n}\right)^i = a^{10/n} \cdot \frac{1 - \left(a^{10/n}\right)^n}{1 - a^{10/n}} = a^{10/n} \cdot \frac{1 - a^{10}}{1 - a^{10/n}}\]

Therefore:

\[S_n = \frac{10}{n} \cdot a^{10/n} \cdot \frac{1 - a^{10}}{1 - a^{10/n}} = \frac{10 \cdot a^{10/n}(1 - a^{10})}{n(1 - a^{10/n})}\]

Step 5: Rewrite the expression

\[S_n = \frac{10(a^{10/n} - a^{10} \cdot a^{10/n})}{n(1 - a^{10/n})} = \frac{10(a^{10/n} - a^{10+10/n})}{n(1 - a^{10/n})}\]

Multiply numerator and denominator by $-1$:

\[S_n = \frac{10(a^{10+10/n} - a^{10/n})}{n(a^{10/n} - 1)}\]

Step 6: Take the limit as $n \to \infty$

Let $h = \frac{10}{n}$, so as $n \to \infty$, we have $h \to 0$:

\[S_n = \frac{10(a^{10+h} - a^h)}{n(a^h - 1)} = \frac{a^h(a^{10} - 1)}{(a^h - 1)/h} \cdot \frac{h}{1}\]

We know that $\lim_{h \to 0} \frac{a^h - 1}{h} = \ln a$, therefore:

\[\lim_{n \to \infty} S_n = \frac{a^0(a^{10} - 1)}{\ln a} = \frac{a^{10} - 1}{\ln a}\]

Verification using antiderivative:

For $a > 0$ and $a \neq 1$:

\[\int_0^{10} a^x \, dx = \left[\frac{a^x}{\ln a}\right]_0^{10} = \frac{a^{10}}{\ln a} - \frac{a^0}{\ln a} = \frac{a^{10} - 1}{\ln a}\]

Final answer:

\[\boxed{\int_0^{10} a^x \, dx = \frac{a^{10} - 1}{\ln a} \quad (a > 0, a \neq 1)}\]