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Hi,
I'm Fiona and this is Tamila 2021 paper 2

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and question one.

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And we're asked to find the value of this
definite integral here,

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which is the integration between one and
four of three square root X + 4 / X ^2 DX.

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So the first thing I'm going to do is to
rewrite this definite integral so that I

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can more clearly see the power that's on
X.

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Now I can see that for this first term,
the power on X is 1/2,

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and for the second term,
the power on X is -2 and now I'm going to

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integrate.

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So now for this first term,
I'm going to be increasing the power on X

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by 1 and then multiplying by the
reciprocal of that power.

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So the power on X is going to be 3 / 2.

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So I'm going to multiply by 2 / 3 times,
3X to the 3 / 2.

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And increasing the power on X here by one
will give me a power of -1.

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So this is going to be -4 X to the power
of -1.

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And I'm evaluating this between 1:00 and
4:00.

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So now I'm going to substitute 4IN where
X is and I end up with 2 * 4 to the power

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of 3 / 2 -,
4 * 4 to the power of -1 and then take

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away.

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And I'm going to input one substitute one
in for X.

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So I get 2 * 1 to the power of 3 / 2 -4 *
1 to the power of -1 and now I'm going to

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have a look and see that 4 to the power
of 3 / 2 is the same as sqrt 4 and then

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raising it to a power of three.

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So sqrt 4 is 2 and 2 ^3 is 8.

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So this is 8 * 2 which is 16 - 4 and 4 to
the power of -1 means 4 / 4.

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So that's going to be a -1 take away one
to any power is just one.

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And so I'm going to be taking away 2 -,
4 and now I have 15 -, -, 2.

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So that's going to give me a value of 17.

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And when I look at the options available
for me to choose from,

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I can see that that is option D which is
the answer to this question.