# What is the area bounded by the curves y=4x^2-25, y=0, x=-1, x=3.5? PS the area is not 54 or 60 I already tried.

Find roots:

So our given area lies below and above the x axis.

It is below in the interval

and above in the interval

So we require the integrals:

1st:

Plugging in upper and lower bounds:

2nd:

Plugging in upper and lower bounds:

Total area

GRAPH:

You probably forgot to remove the negation when adding the two areas.

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To find the area bounded by the curves y = 4x^2 - 25, y = 0, x = -1, and x = 3.5, you need to integrate the function (4x^2 - 25) between the given limits of integration, -1 and 3.5, and then take the absolute value of the result. This is because the curve may dip below the x-axis.

So, the integral you need to solve is:

[ A = \left| \int_{-1}^{3.5} (4x^2 - 25) , dx \right| ]

[ A = \left| \left[ \frac{4}{3}x^3 - 25x \right]_{-1}^{3.5} \right| ]

[ A = \left| \left( \frac{4}{3}(3.5)^3 - 25(3.5) \right) - \left( \frac{4}{3}(-1)^3 - 25(-1) \right) \right| ]

[ A = \left| \left( \frac{4}{3}(42.875) - 87.5 \right) - \left( \frac{4}{3}(-1) + 25 \right) \right| ]

[ A = \left| \left( 57.166 - 87.5 \right) - \left( -\frac{4}{3} + 25 \right) \right| ]

[ A = \left| \left( -30.334 \right) - \left( 23.666 \right) \right| ]

[ A = \left| -54 \right| ]

So, the area bounded by the curves (y = 4x^2 - 25), (y = 0), (x = -1), and (x = 3.5) is (54).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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