Sketch the region whose area is given by the definite integral and then use a geometric formula to evaluate the integral.Round to one decimal place if necessary.
A)-26.5
B)6
C)44
D)105.5
E)-30.5

Question

Sketch the region whose area is given by the definite integral and then use a geometric formula to evaluate the integral.Round to one decimal place if necessary. ​  ​
A)-26.5 
B)6 
C)44 
D)105.5 
E)-30.5

Quizplus · Accepted Answer

Since the area is positive, the region must be above the x-axis. The integrand is $2y-1$, which means the region is bounded by the lines $y=\frac{1}{2}$ and $y=3$ and the $y$-axis. Sketching this region, we see that it is a trapezoid:

[asy]
unitsize(.5cm);

fill((0,0)--(0,3)--(2,5)--(2,0)--cycle, lightgray);
draw((0,-1)--(0,6), Arrow);
draw((-1,0)--(3,0), Arrow);
label("$y$", (0,6), N);
label("$x$", (3,0), E);
label("$\frac{1}{2}$", (-.3,.5));
draw((-.2,.5)--(.2,.5));
label("$3$",(2.5,3));
draw((2.8,3)--(2.6,3));
draw((0,0)--(0,3)--(2,5)--(2,0)--cycle, linewidth(1));
draw((2,5)--(2,6), linetype("4 4"));

[/asy]

To evaluate the integral, we need to integrate with respect to $y$. At height $y$, the width of the trapezoid is $2y-1$ (since the $y$-coordinate of the point where the line $x=2$ intersects the trapezoid is $2y-1$). Thus, integrating with respect to $y$, we have \begin{align*}
\int_{1/2}^3 (2y-1) dy &= \left[y^2-yight]_{1/2}^3\
&= \left[ 9 - 3 - \frac{1}{4} + \frac{1}{2}ight] \
&= \boxed{44}.
\end{align*}

Sketch the Region Whose Area Is Given by the Definite