Question

I cannot solve this. can u plis solve this huhuhuhuhuhuhuhuhuhu

Answer 1

Not middle school.

$f(x) = \frac 1 8 ( e^{4x} + 3 )$

We're after the integral from 0 to B, less the right triangle OAB. So we need to find A and B.

$A = f(0) = \frac 1 8(1 + 3) = \frac 1 2$

The slope at 0 is $f'(0)$

$f'(x) = \frac 1 8 ( 4 e^{4x} ) = \frac 1 2 e^{4x}$

$f'(0) = \frac 1 8(4) = \frac 1 2$

The normal has negative reciprocal slope, so m=-2 through (0,1/2)

$y - \frac 1 2 = -2 x$

The x intercept (when y=0) is B:

$-\frac 1 2 = -2B$

$B = \frac 1 4$

The right triangle area is

$\frac 1 2 AB = \frac{1}{16}$

The integral is

$\displaystyle \int_0^{\frac 1 4} \frac 1 8 ( e^{4x} + 3 ) dx = \frac 1 8 ( \frac 1 4 e^{4x} + 3x ) |_0^{\frac 1 4}$

$= \frac 1 8( (e/4 + 3/4) - 1/4) = \frac{1}{32}(e+2)$

The area we seek is the difference,

$A= \frac{1}{32}(e+2) - \frac {1}{16} = \dfrac{e}{32}$

Answer: e/32