Let f (x) = $2 ^ { 6 x }$ on the interval [0,1]. Let the interval be divided into three equal subintervals. Find the value of the Riemann sum $\sum _ { i = 1 } ^ { 3 } f \left( x _ { i } ^ { * } \right) \Delta x _ { i }$ if each $x _ { i } ^ { * }$ is the left endpoint of its subinterval.
A) $\frac { 13 } { 3 }$
B)21
C)7
D)28
E)8
F)42
G)14
H)84

Question

Let f (x) = $2 ^ { 6 x }$ on the interval [0,1]. Let the interval be divided into three equal subintervals. Find the value of the Riemann sum $\sum _ { i = 1 } ^ { 3 } f \left( x _ { i } ^ { * } \right) \Delta x _ { i }$ if each $x _ { i } ^ { * }$ is the left endpoint of its subinterval.
A) $\frac { 13 } { 3 }$ 
B)21
C)7
D)28
E)8
F)42
G)14
H)84

Quizplus · Accepted Answer

The interval [0,1] is divided into three equal subintervals: [0, 1/3], [1/3, 2/3], and [2/3, 1]. The left endpoints are 0, 1/3, and 2/3. The width of each subinterval, $\Delta x$, is 1/3. The Riemann sum is $f(0) \cdot \frac{1}{3} + f(\frac{1}{3}) \cdot \frac{1}{3} + f(\frac{2}{3}) \cdot \frac{1}{3}$. Calculating each term: $f(0) = 2^{6 \cdot 0} = 1$, $f(\frac{1}{3}) = 2^{6 \cdot \frac{1}{3}} = 2^2 = 4$, $f(\frac{2}{3}) = 2^{6 \cdot \frac{2}{3}} = 2^4 = 16$. Thus, the sum is $(1 \cdot \frac{1}{3}) + (4 \cdot \frac{1}{3}) + (16 \cdot \frac{1}{3}) = \frac{1}{3} + \frac{4}{3} + \frac{16}{3} = \frac{21}{3} = 7$.

Let F (X) = 26x2 ^ { 6 x }26x On the Interval [0,1]

Let F (X) = $2 ^ { 6 x }$ On the Interval [0,1]