Not middle school.
[tex]f(x) = \frac 1 8 ( e^{4x} + 3 )[/tex]
We're after the integral from 0 to B, less the right triangle OAB. So we need to find A and B.
[tex]A = f(0) = \frac 1 8(1 + 3) = \frac 1 2[/tex]
The slope at 0 is [tex]f'(0)[/tex]
[tex]f'(x) = \frac 1 8 ( 4 e^{4x} ) = \frac 1 2 e^{4x}[/tex]
[tex]f'(0) = \frac 1 8(4) = \frac 1 2[/tex]
The normal has negative reciprocal slope, so m=-2 through (0,1/2)
[tex]y - \frac 1 2 = -2 x[/tex]
The x intercept (when y=0) is B:
[tex]-\frac 1 2 = -2B [/tex]
[tex]B = \frac 1 4[/tex]
The right triangle area is
[tex]\frac 1 2 AB = \frac{1}{16}[/tex]
The integral is
[tex]\displaystyle \int_0^{\frac 1 4} \frac 1 8 ( e^{4x} + 3 ) dx = \frac 1 8 ( \frac 1 4 e^{4x} + 3x ) |_0^{\frac 1 4}[/tex]
[tex] = \frac 1 8( (e/4 + 3/4) - 1/4) = \frac{1}{32}(e+2)[/tex]
The area we seek is the difference,
[tex] A= \frac{1}{32}(e+2) - \frac {1}{16} = \dfrac{e}{32}[/tex]
Answer: e/32
