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You are watching: Find the area of the region bounded by the graphs of the equations We are asked to uncover the area that the an ar bounded through `f(x)=-x^2+4x` and also `y=0` .

We can element the quadratic come y=-x(x-4); the x-intercepts space 0 and 4.

The equivalent question then is to discover the identify integral `int_0^4 (-x^2+4x)dx` .

`int_0^4 (-x^2+4x)dx=-x^3/3+2x^2 |_0^4`

`=(-64/3+32)-(0+0)=32/3 `

for this reason the area...

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We room asked to discover the area the the an ar bounded through `f(x)=-x^2+4x` and `y=0` .

We can aspect the quadratic come y=-x(x-4); the x-intercepts are 0 and 4.

The identical question then is to find the definite integral `int_0^4 (-x^2+4x)dx` .

`int_0^4 (-x^2+4x)dx=-x^3/3+2x^2 |_0^4`

`=(-64/3+32)-(0+0)=32/3 `

So the area bounded by the offered curve and also the x-axis is 32/3.

See more: A Relationship Can Be Damaged By, Can A Damaged Relationship Be Repaired

An different approach is to use a Riemann sum:

The area can be uncovered by `lim_n->oo sum_i=1^n f(c_i)Delta x_i`

If we use a constant partition and also choose the right-hand endpoint in each subinterval us get:

`A=lim_n->oo sum_i=1^n (-((4i)/n)^2+4((4i)/n))(4/n) `

`=lim_n->oo<(-64)/(n^2)sum_i=1^n(i^2/n-i)> `

`=lim_n->oo<(-64)/n^2(1/n((n(n+1)(2n+1))/6)-(n(n+1))/2)> `

`=lim_n->oo<(-64)/n^2(1/3n^2+1/2n+1/3-n^2/2-n/2)> `

`=lim_n->oo<(-64)/3-(32)/n-(64)/n^3+32+(32)/n> `

`=(-64)/3+32=32/3=10.bar(6) `

The graph:

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