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What is the area of the region between the graphs of
f(x)=x^(2)+12 x and
g(x)=3x^(2)+10 from
x=1 to
x=4 ?
Choose 1 answer:
(A) 77
(B)
(64)/(3)
(C) 18
(D) 45

What is the area of the region between the graphs of $f(x)=x^{2}+12 x$ and $g(x)=3 x^{2}+10$ from $x=1$ to $x=4$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $77$ $\newline$ (B) $\frac{64}{3}$ $\newline$ (C) $18$ $\newline$ (D) $45$

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Find equations of tangent lines using limits

Full solution

Q. What is the area of the region between the graphs of

f(x)=x^{2}+12 x

and

g(x)=3 x^{2}+10

from

x=1

to

x=4

?

\newline

Choose

1

answer:

\newline

(A)

77

\newline

(B)

\frac{64}{3}

\newline

(C)

18

\newline

(D)

45

Set up integral: First, we need to set up the integral to find the area between the two curves. The area $A$ can be found by integrating the difference between the functions $f(x)$ and $g(x)$ from $x=1$ to $x=4$ . $A = \int_{x=1}^{x=4} (f(x) - g(x)) \, dx$
Write functions: Now we need to write down the functions $f(x)$ and $g(x)$ and find the difference $f(x) - g(x)$ .
$f(x) = x^2 + 12x$
$g(x) = 3x^2 + 10$
The difference is:
$f(x) - g(x) = (x^2 + 12x) - (3x^2 + 10) = -2x^2 + 12x - 10$
Integrate difference: Next, we integrate the function $-2x^2 + 12x - 10$ from $x=1$ to $x=4$ .
A = $\int_{x=1}^{x=4} (-2x^2 + 12x - 10) \, dx$
To integrate, we find the antiderivative of $-2x^2 + 12x - 10$ .
Antiderivative: $(-2/3)x^3 + 6x^2 - 10x$
Evaluate antiderivative: We now evaluate the antiderivative from $x=1$ to $x=4$ .
$A = \left[\left(-\frac{2}{3}\right)x^3 + 6x^2 - 10x\right]$ from $x=1$ to $x=4$
$A = \left[\left(-\frac{2}{3}\right)(4)^3 + 6(4)^2 - 10(4)\right] - \left[\left(-\frac{2}{3}\right)(1)^3 + 6(1)^2 - 10(1)\right]$
Calculate values: Calculate the values at $x=4$ and $x=1$ .
$A = [(-\frac{2}{3})(64) + 6(16) - 40] - [(-\frac{2}{3})(1) + 6(1) - 10]$
$A = [(-\frac{128}{3}) + 96 - 40] - [(-\frac{2}{3}) + 6 - 10]$
$A = [(-\frac{128}{3}) + 56] - [(-\frac{2}{3}) - 4]$
Simplify expression: Simplify the expression. $\newline$ $A = \frac{{(-128 + 168)}}{3} - \frac{{(-2 - 12)}}{3}$ $\newline$ $A = \frac{{40}}{3} - \frac{{-14}}{3}$ $\newline$ $A = \frac{{40}}{3} + \frac{{14}}{3}$ $\newline$ $A = \frac{{54}}{3}$ $\newline$ $A = 18$

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