Find the area of the region under the graph of the function
On the interval , using the fundamental theorem of calculus. Then verify your result using geometry.

Question

Find the area of the region under the graph of the function ​  ​
On the interval  , using the fundamental theorem of calculus. Then verify your result using geometry.
​

Quizplus · Accepted Answer

Using the fundamental theorem of calculus, we can find the area under the graph of the function by taking the antiderivative of the function and evaluating it at the endpoints of the interval:

$$\int_{-2}^{1} (2x+2) dx = \left[x^2 + 2x\right]_{-2}^{1} = (1^2+2\cdot1)-((-2)^2+2\cdot(-2)) = 3+8 = 11$$

So the area under the graph is 11 sq units.

To verify this result using geometry, we can graph the function and the interval and find the area of the corresponding trapezoid:

[asy]
unitsize(0.45 cm);

pair[] A,B;
real f(real x)
{
  return((x^2+2*x-7)/2);
}

A[0] = (-2,0);
B[0] = (1,0);
A[1] = A[0];
B[1] = (A[1].x,f(A[1].x));
A[2] = B[0];
B[2] = (B[0].x,f(B[0].x));
A[3] = B[1];
B[3] = B[2];

fill(B[1]--B[2]--(B[2].x,0)--(B[1].x,0)--cycle,gray(0.7));
draw(A[0]--B[0],dashed);
draw(B[0]--B[3]);
draw(A[1]--B[1]);
draw(A[2]--B[2]);
draw((-2.5,0)--(2.5,0));
draw((0,-2)--(0,16));
label("$y = \frac{1}{2}(x^2+2x-7)$", (2,7));
label("$(-2,0)$", A[0], S);
label("$(1,0)$", B[0], S);
[/asy]

The height of the trapezoid is the maximum value of the function on the interval, which occurs at $x = -1$, where $y = \frac{1}{2}((-1)^2+2(-1)-7) = -\frac{5}{2}$. The lengths of the top and bottom sides are $f(-2) = \frac{1}{2}((-2)^2+2(-2)-7) = -\frac{9}{2}$ and $f(1) = \frac{1}{2}(1^2+2(1)-7) = -\frac{3}{2}$, respectively. Therefore, the area of the trapezoid is:

$$\frac{1}{2} \cdot (-\frac{9}{2} - \frac{3}{2}) \cdot (-\frac{5}{2}) = \frac{12}{4} \cdot \frac{5}{2} = 15$$

This is not equal to the value obtained using the fundamental theorem of calculus, so we made an error in our geometry calculation. To find the correct area using geometry, we can divide the trapezoid into a rectangle and a right triangle:

[asy]
unitsize(0.45 cm);

draw((-2.5,0)--(2.5,0));
draw((0,-2)--(0,16));
label("$y = \frac{1}{2}(x^2+2x-7)$", (2,7));

fill((-2,0)--(-2,-7)--(1,-2)--(1,0)--cycle,gray(0.7));
draw((-2,-7)--(-2,0)--(1,0)--(1,-2));
draw((-2,-7)--(1,-7),dashed);
draw((1,-2)--(2,-2),dashed);
draw((1,-2)--(1,0),dashed);
draw(rightanglemark((1,-2),(1,0),(2,0),10));

label("$(-2,0)$", (-2,0), S);
label("$(1,0)$", (1,0), S);
label("$(1,-2)$", (1,-2), E);
label("$(-2,-7)$", (-2,-7), W);
label("$(2,-2)$", (2,-2), E);
[/asy]

The area of the rectangle is $(1-(-2))\cdot(-2) = 6$, and the area of the right triangle is $\frac{1}{2} \cdot (1-(-2)) \cdot (-7-(-2)) = \frac{15}{2}$. Therefore, the total area under the graph is:

$$6 + \frac{15}{2} = \frac{12+15}{2} = \frac{27}{2} = 13.5 \approx 14$$

This is closest to option C, 12 sq units. However, none of the answer choices match the exact value of either the calculus or geometry calculation.

Find the Area of the Region Under the Graph of the Function