Find the Area Using Limits - Pre-Calculus

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Question

What is the area under the curve of the function

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

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

Answer

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

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

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

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

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

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

Compare your answer with the correct one above

Question

What is the area under the curve of the function

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

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

Answer

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

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

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

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

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

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

Compare your answer with the correct one above

Question

What is the area under the curve of the function

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

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

Answer

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

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

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

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

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

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

Compare your answer with the correct one above

Question

What is the area under the curve of the function

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

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

Answer

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

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

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

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

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

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

Compare your answer with the correct one above

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