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What is the area under the graph of #f(x) = x^2 + 2# on #[1, 2]#?

Answer 1

Emma Aldridge

#13/3# square units.

The bounds of integration are given, therefore we can write our expression for area right away.

#A = int_1^2 x^2 + 2#

#A = [1/3x^3 + 2x]_1^2#

#A = 1/3(2)^3 + 2(2) - (1/3(1)^3 + 2(1))#

#A = 8/3 + 4 - 1/3 - 2#

#A = 7/3 + 2#

#A = 13/3#

Hopefully this helps!

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Answer 2

Cameron Alexander

To find the area under the graph of ( f(x) = x^2 + 2 ) on the interval ([1, 2]), we need to compute the definite integral of ( f(x) ) over the interval ([1, 2]). Using the definite integral formula, we have:

[ \text{Area} = \int_{1}^{2} (x^2 + 2) , dx ]

Evaluating this integral gives:

[ \text{Area} = \left[ \frac{x^3}{3} + 2x \right]_{1}^{2} ] [ = \left( \frac{2^3}{3} + 2(2) \right) - \left( \frac{1^3}{3} + 2(1) \right) ] [ = \left( \frac{8}{3} + 4 \right) - \left( \frac{1}{3} + 2 \right) ] [ = \left( \frac{8}{3} + \frac{12}{3} \right) - \left( \frac{1}{3} + \frac{6}{3} \right) ] [ = \left( \frac{20}{3} \right) - \left( \frac{7}{3} \right) ] [ = \frac{13}{3} ]

So, the area under the graph of ( f(x) = x^2 + 2 ) on ([1, 2]) is ( \frac{13}{3} ).

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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.