What is the net area between #f(x) = x-xsqrt(4x-3) # and the x-axis over #x in [1, 4 ]#?

Answer 1

#A=int_1^4[0-(x-xsqrt(4x-3))]*dx=int_1^4-x+xsqrt(4x-3)*dx=13.44 (unite)^2#

show below:

the sketch of the function #f(x) = x-xsqrt(4x-3)#
graph{x-xsqrt(4x-3) [-10, 10, -5, 5]}

the area wanted is between x=1 and x=4 as shown in the graph below

now , set up the integral

#A=int_1^4[0-(x-xsqrt(4x-3))]*dx=int_1^4-x+xsqrt(4x-3)*dx#

#=[(4*x-3)^(5/2)/40+(4*x-3)^(3/2)/8-x^2/2]_1^4=[(9*13^(3/2)-153)/20]=13.44(unite)^2#

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Answer 2

To find the net area between the function ( f(x) = x - x\sqrt{4x - 3} ) and the x-axis over the interval ( x ) in ([1, 4]), we need to compute the definite integral of ( |f(x)| ) over this interval.

The function ( f(x) = x - x\sqrt{4x - 3} ) intersects the x-axis where ( f(x) = 0 ). So, we solve for ( x ) when ( f(x) = 0 ):

[ x - x\sqrt{4x - 3} = 0 ]

[ x(1 - \sqrt{4x - 3}) = 0 ]

Either ( x = 0 ) or ( 1 - \sqrt{4x - 3} = 0 ). Since we're only interested in the interval ([1, 4]), we consider the latter equation:

[ 1 - \sqrt{4x - 3} = 0 ]

[ \sqrt{4x - 3} = 1 ]

[ 4x - 3 = 1 ]

[ 4x = 4 ]

[ x = 1 ]

So, the function crosses the x-axis at ( x = 1 ).

To find the area over the interval ([1, 4]), we integrate ( |f(x)| ) from ( x = 1 ) to ( x = 4 ):

[ \text{Net Area} = \int_{1}^{4} |f(x)| , dx ]

[ = \int_{1}^{4} |x - x\sqrt{4x - 3}| , dx ]

[ = \int_{1}^{4} x - x\sqrt{4x - 3} , dx ]

Now, we can find the integral of each term separately and subtract:

[ \text{Net Area} = \left[\frac{x^2}{2} - \frac{2}{3}(4x - 3)^{\frac{3}{2}}\right]_{1}^{4} ]

[ = \left[\frac{x^2}{2} - \frac{2}{3}(4x - 3)^{\frac{3}{2}}\right]_{1}^{4} ]

[ = \left[\frac{x^2}{2} - \frac{2}{3}(64 - 3)^{\frac{3}{2}} - \left(\frac{1^2}{2} - \frac{2}{3}(4 - 3)^{\frac{3}{2}}\right)\right] ]

[ = \left[\frac{16}{2} - \frac{2}{3}(61)^{\frac{3}{2}} - \left(\frac{1}{2} - \frac{2}{3}(1)^{\frac{3}{2}}\right)\right] ]

[ = \left[8 - \frac{2}{3}(61)^{\frac{3}{2}} - \left(\frac{1}{2} - \frac{2}{3}\right)\right] ]

[ = 8 - \frac{2}{3}(61)^{\frac{3}{2}} - \frac{1}{2} + \frac{2}{3} ]

[ ≈ 8 - 27.99 - 0.5 + \frac{2}{3} ]

[ ≈ -20.49 ]

So, the net area between ( f(x) = x - x\sqrt{4x - 3} ) and the x-axis over the interval ( x ) in ([1, 4]) is approximately ( -20.49 ) square units.

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Answer from HIX Tutor

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.

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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