# What is the area under #f(x)=x^2-3x+5# in #x in[0,1] #?

Drawing the graph of the function we get

graph{x^2-3x+5 [-9.14, 19.34, -3.53, 10.71]}

Since there are no x-intercepts or turning points between the interval [0,1], we may integrate throughout the interval to obtain the required area as

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To find the area under the curve (f(x) = x^2 - 3x + 5) in the interval ([0, 1]), you need to compute the definite integral of (f(x)) with respect to (x) over that interval. Using the fundamental theorem of calculus, integrate (f(x)) from 0 to 1:

[ \int_{0}^{1} (x^2 - 3x + 5) , dx ]

[= \left[\frac{x^3}{3} - \frac{3x^2}{2} + 5x\right]_{0}^{1}]

[= \left(\frac{1}{3} - \frac{3}{2} + 5\right) - \left(0 - 0 + 0\right)]

[= \left(\frac{1}{3} - \frac{3}{2} + 5\right)]

[= \left(\frac{1}{3} - \frac{9}{6} + \frac{30}{6}\right)]

[= \left(\frac{1}{3} - \frac{9}{6} + \frac{30}{6}\right)]

[= \left(\frac{1}{3} + \frac{21}{6}\right)]

[= \left(\frac{1}{3} + \frac{7}{2}\right)]

[= \left(\frac{1}{3} + \frac{21}{6}\right)]

[= \left(\frac{1}{3} + \frac{7}{2}\right)]

[= \frac{1}{3} + \frac{21}{6}]

[= \frac{1}{3} + \frac{7}{2}]

[= \frac{1}{3} + \frac{21}{6}]

[= \frac{1}{3} + \frac{7}{2}]

[= \frac{1 + 21}{3 + 6}]

[= \frac{22}{9}]

Therefore, the area under the curve (f(x) = x^2 - 3x + 5) in the interval ([0, 1]) is (\frac{22}{9}) square units.

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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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