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Precalculus : Find the Area Using Limits

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Example Question #1 : Area Under A Curve

What is the area under the curve of the function

$\small f(x)=3(x-1)^2+1$ $\displaystyle \small f(x)=3(x-1)^2+1$

from $\small x=0$ $\displaystyle \small x=0$ to $\small x=2$ $\displaystyle \small x=2$ .

Possible Answers:

$\small 3$ $\displaystyle \small 3$

$\small 3.5$ $\displaystyle \small 3.5$

$\small 2$ $\displaystyle \small 2$

$\small 4$ $\displaystyle \small 4$

$\displaystyle 6$

Correct answer:

$\small 4$ $\displaystyle \small 4$

Explanation:

The area under the curve of the function $\small f(x)$ $\displaystyle \small f(x)$ is the definite integral from $\small x=0$ $\displaystyle \small x=0$ to $\small x=2$ $\displaystyle \small x=2$ .

Remember when integrating, we will increase the exponent by one and then divide the whole term by the value of the new exponent.

$\small \int_0^2 f(x)dx=\int_0^2 [3(x-1)^2+1]dx =(x-1)^3+x]_0^2$ $\displaystyle \small \int_0^2 f(x)dx=\int_0^2 [3(x-1)^2+1]dx =(x-1)^3+x]_0^2$

From here, we find the difference between the function values of the boundaries.

$\small \small \small =[(2-1)^3+2]-[(0-1)^3+0]=1^3+2-(-1)^3$ $\displaystyle \small \small \small =[(2-1)^3+2]-[(0-1)^3+0]=1^3+2-(-1)^3$

$\small =1+2+1=4$ $\displaystyle \small =1+2+1=4$

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Precalculus : Find the Area Using Limits

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Example Question #1 : Area Under A Curve

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