Find the Area Using Limits - Pre-Calculus
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What is the area under the curve of the function

from
to
.
What is the area under the curve of the function
from to
.
Tap to see back →
The area under the curve of the function
is the definite integral from
to
.
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]_$0^2$](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/348947/gif.latex)
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$](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/348948/gif.latex)

The area under the curve of the function is the definite integral from
to
.
Remember when integrating, we will increase the exponent by one and then divide the whole term by the value of the new exponent.
From here, we find the difference between the function values of the boundaries.
What is the area under the curve of the function

from
to
.
What is the area under the curve of the function
from to
.
Tap to see back →
The area under the curve of the function
is the definite integral from
to
.
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]_$0^2$](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/348947/gif.latex)
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$](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/348948/gif.latex)

The area under the curve of the function is the definite integral from
to
.
Remember when integrating, we will increase the exponent by one and then divide the whole term by the value of the new exponent.
From here, we find the difference between the function values of the boundaries.
What is the area under the curve of the function

from
to
.
What is the area under the curve of the function
from to
.
Tap to see back →
The area under the curve of the function
is the definite integral from
to
.
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]_$0^2$](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/348947/gif.latex)
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$](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/348948/gif.latex)

The area under the curve of the function is the definite integral from
to
.
Remember when integrating, we will increase the exponent by one and then divide the whole term by the value of the new exponent.
From here, we find the difference between the function values of the boundaries.
What is the area under the curve of the function

from
to
.
What is the area under the curve of the function
from to
.
Tap to see back →
The area under the curve of the function
is the definite integral from
to
.
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]_$0^2$](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/348947/gif.latex)
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$](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/348948/gif.latex)

The area under the curve of the function is the definite integral from
to
.
Remember when integrating, we will increase the exponent by one and then divide the whole term by the value of the new exponent.
From here, we find the difference between the function values of the boundaries.