$\text { Find the area of the region bounded by the graphs of } y = \frac { 16 } { x ^ { 2 } + 5 x + 6 } , y = 0 , x = 0 \text {, }$ and $x = 1$.
A)
$A = 16 \log \left( \frac { 9 } { 8 } \right)$
B)
$A = 16 \log \left( \frac { 8 } { 9 } \right)$
C)
$A = 16 + \log \left( \frac { 5 } { 8 } \right)$
D)
$A = 16 \log \left( \frac { 5 } { 6 } \right)$
E)
$A = 16 - \log \left( \frac { 5 } { 9 } \right)$

Question

$\text { Find the area of the region bounded by the graphs of } y = \frac { 16 } { x ^ { 2 } + 5 x + 6 } , y = 0 , x = 0 \text {, }$ and $x = 1$.
A) 
$A = 16 \log \left( \frac { 9 } { 8 } \right)$
B)
$A = 16 \log \left( \frac { 8 } { 9 } \right)$
C)
$A = 16 + \log \left( \frac { 5 } { 8 } \right)$
D)
$A = 16 \log \left( \frac { 5 } { 6 } \right)$
E)
$A = 16 - \log \left( \frac { 5 } { 9 } \right)$

Quizplus · Accepted Answer

The area under the curve $y = \frac{16}{x^2 + 5x + 6}$ from $x = 0$ to $x = 1$ can be found by integrating the function with respect to $x$ over the interval $[0, 1]$. The denominator $x^2 + 5x + 6$ can be factored into $(x+2)(x+3)$, making the integral easier to solve using partial fractions. The result of this integration leads to $16 \log \left( \frac{9}{8} \right)$, which matches choice A.

Find the area of the region bounded by the graphs of y=16x2+5x+6,y=0,x=0, \text { Find the area of the region bounded by the graphs of } y = \frac { 16 } { x ^ { 2 } + 5 x + 6 } , y = 0 , x = 0 \text {, } Find the area of the region bounded by the graphs of y=x2+5x+616​,y=0,x=0,

$\text { Find the area of the region bounded by the graphs of } y = \frac { 16 } { x ^ { 2 } + 5 x + 6 } , y = 0 , x = 0 \text {, }$