Area between the curve $y$, the $x$-axis and the lines
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Find the total area between the curve $y=x^2-4x+3$, the $x$-axis and the lines $x=0$ and $x=3$.
I have drawn the graph and concluded that:
$$int_0^1 0 - (x^2-4x+3), dx + int_1^3 0 - (x^2-4x+3), dx = -frac-263$$ Which differs from the answer key: $4$. What is the mistake here? Is it feasiable to solve this problem without relying on the graph?
calculus integration definite-integrals area
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up vote
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favorite
Find the total area between the curve $y=x^2-4x+3$, the $x$-axis and the lines $x=0$ and $x=3$.
I have drawn the graph and concluded that:
$$int_0^1 0 - (x^2-4x+3), dx + int_1^3 0 - (x^2-4x+3), dx = -frac-263$$ Which differs from the answer key: $4$. What is the mistake here? Is it feasiable to solve this problem without relying on the graph?
calculus integration definite-integrals area
the first integral should be positive, because the upper graphic is $x^2-4x+3$
â haqnatural
Aug 31 at 4:04
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up vote
1
down vote
favorite
up vote
1
down vote
favorite
Find the total area between the curve $y=x^2-4x+3$, the $x$-axis and the lines $x=0$ and $x=3$.
I have drawn the graph and concluded that:
$$int_0^1 0 - (x^2-4x+3), dx + int_1^3 0 - (x^2-4x+3), dx = -frac-263$$ Which differs from the answer key: $4$. What is the mistake here? Is it feasiable to solve this problem without relying on the graph?
calculus integration definite-integrals area
Find the total area between the curve $y=x^2-4x+3$, the $x$-axis and the lines $x=0$ and $x=3$.
I have drawn the graph and concluded that:
$$int_0^1 0 - (x^2-4x+3), dx + int_1^3 0 - (x^2-4x+3), dx = -frac-263$$ Which differs from the answer key: $4$. What is the mistake here? Is it feasiable to solve this problem without relying on the graph?
calculus integration definite-integrals area
calculus integration definite-integrals area
asked Aug 31 at 3:56
Mauricio Mendes
968
968
the first integral should be positive, because the upper graphic is $x^2-4x+3$
â haqnatural
Aug 31 at 4:04
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the first integral should be positive, because the upper graphic is $x^2-4x+3$
â haqnatural
Aug 31 at 4:04
the first integral should be positive, because the upper graphic is $x^2-4x+3$
â haqnatural
Aug 31 at 4:04
the first integral should be positive, because the upper graphic is $x^2-4x+3$
â haqnatural
Aug 31 at 4:04
add a comment |Â
3 Answers
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The required area is the green area:
$hspace3cm$
The proper set up of integrals:
$$int_0^1 (x^2-4x+3)-0, dx + int_1^3 0 - (x^2-4x+3), dx = frac83.$$
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$int_0^1 (x^2-4x+3)mathbb dx-int_1^3(x^2-4x+3)mathbb dx=[(fracx^33-2x^2+3x)]_0^1-[fracx^33-2x^2+3x]_1^3=frac43-[0-frac43]=frac83$
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Hint:
Area under curve of $y=f(x)$ in the domain of $a < x < b$ is given by
beginalign
mathrmArea ; = ; int_a^b Big|, f(x) , Big| ; dx
endalign
Take absolute value of your integrand.
For your case, $f(x) = x^2 - 4x + 3$.
Factorization gives $f(x) = (x-3)(x-1)$.
Now you can easily see when $f(x) > 0$ or $<0$ without using a graph.
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The required area is the green area:
$hspace3cm$
The proper set up of integrals:
$$int_0^1 (x^2-4x+3)-0, dx + int_1^3 0 - (x^2-4x+3), dx = frac83.$$
add a comment |Â
up vote
1
down vote
accepted
The required area is the green area:
$hspace3cm$
The proper set up of integrals:
$$int_0^1 (x^2-4x+3)-0, dx + int_1^3 0 - (x^2-4x+3), dx = frac83.$$
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The required area is the green area:
$hspace3cm$
The proper set up of integrals:
$$int_0^1 (x^2-4x+3)-0, dx + int_1^3 0 - (x^2-4x+3), dx = frac83.$$
The required area is the green area:
$hspace3cm$
The proper set up of integrals:
$$int_0^1 (x^2-4x+3)-0, dx + int_1^3 0 - (x^2-4x+3), dx = frac83.$$
answered Aug 31 at 4:37
farruhota
15.2k2734
15.2k2734
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add a comment |Â
up vote
1
down vote
$int_0^1 (x^2-4x+3)mathbb dx-int_1^3(x^2-4x+3)mathbb dx=[(fracx^33-2x^2+3x)]_0^1-[fracx^33-2x^2+3x]_1^3=frac43-[0-frac43]=frac83$
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up vote
1
down vote
$int_0^1 (x^2-4x+3)mathbb dx-int_1^3(x^2-4x+3)mathbb dx=[(fracx^33-2x^2+3x)]_0^1-[fracx^33-2x^2+3x]_1^3=frac43-[0-frac43]=frac83$
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$int_0^1 (x^2-4x+3)mathbb dx-int_1^3(x^2-4x+3)mathbb dx=[(fracx^33-2x^2+3x)]_0^1-[fracx^33-2x^2+3x]_1^3=frac43-[0-frac43]=frac83$
$int_0^1 (x^2-4x+3)mathbb dx-int_1^3(x^2-4x+3)mathbb dx=[(fracx^33-2x^2+3x)]_0^1-[fracx^33-2x^2+3x]_1^3=frac43-[0-frac43]=frac83$
answered Aug 31 at 4:22
Chris Custer
6,3472622
6,3472622
add a comment |Â
add a comment |Â
up vote
0
down vote
Hint:
Area under curve of $y=f(x)$ in the domain of $a < x < b$ is given by
beginalign
mathrmArea ; = ; int_a^b Big|, f(x) , Big| ; dx
endalign
Take absolute value of your integrand.
For your case, $f(x) = x^2 - 4x + 3$.
Factorization gives $f(x) = (x-3)(x-1)$.
Now you can easily see when $f(x) > 0$ or $<0$ without using a graph.
add a comment |Â
up vote
0
down vote
Hint:
Area under curve of $y=f(x)$ in the domain of $a < x < b$ is given by
beginalign
mathrmArea ; = ; int_a^b Big|, f(x) , Big| ; dx
endalign
Take absolute value of your integrand.
For your case, $f(x) = x^2 - 4x + 3$.
Factorization gives $f(x) = (x-3)(x-1)$.
Now you can easily see when $f(x) > 0$ or $<0$ without using a graph.
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Hint:
Area under curve of $y=f(x)$ in the domain of $a < x < b$ is given by
beginalign
mathrmArea ; = ; int_a^b Big|, f(x) , Big| ; dx
endalign
Take absolute value of your integrand.
For your case, $f(x) = x^2 - 4x + 3$.
Factorization gives $f(x) = (x-3)(x-1)$.
Now you can easily see when $f(x) > 0$ or $<0$ without using a graph.
Hint:
Area under curve of $y=f(x)$ in the domain of $a < x < b$ is given by
beginalign
mathrmArea ; = ; int_a^b Big|, f(x) , Big| ; dx
endalign
Take absolute value of your integrand.
For your case, $f(x) = x^2 - 4x + 3$.
Factorization gives $f(x) = (x-3)(x-1)$.
Now you can easily see when $f(x) > 0$ or $<0$ without using a graph.
answered Aug 31 at 4:06
K_inverse
22829
22829
add a comment |Â
add a comment |Â
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the first integral should be positive, because the upper graphic is $x^2-4x+3$
â haqnatural
Aug 31 at 4:04