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Use integration by substitution to show that $$\int_{-4}^{6} \sqrt{4x+1} \: dx = \frac{875}{12}$$ Fully justify your answer. - AQA - A-Level Maths Pure - Question 5 - 2020 - Paper 2

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Use-integration-by-substitution-to-show-that--$$\int_{-4}^{6}-\sqrt{4x+1}-\:-dx-=-\frac{875}{12}$$--Fully-justify-your-answer.-AQA-A-Level Maths Pure-Question 5-2020-Paper 2.png

Use integration by substitution to show that $$\int_{-4}^{6} \sqrt{4x+1} \: dx = \frac{875}{12}$$ Fully justify your answer.

Worked Solution & Example Answer:Use integration by substitution to show that $$\int_{-4}^{6} \sqrt{4x+1} \: dx = \frac{875}{12}$$ Fully justify your answer. - AQA - A-Level Maths Pure - Question 5 - 2020 - Paper 2

Step 1

Use a suitable substitution

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Answer

Let us choose the substitution:

u=4x+1.u = 4x + 1.
Then, the differential is:

du=4dx,ordx=du4.du = 4 \: dx, \quad \text{or} \quad dx = \frac{du}{4}.
We also need to change the limits of integration:

When x=4x = -4, u=4(4)+1=15u = 4(-4) + 1 = -15;
When x=6x = 6, u=4(6)+1=25.u = 4(6) + 1 = 25.

Step 2

Completes substitution to obtain correct integrand

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Answer

Substituting into the integral, we get:

$$\int_{-15}^{25} \sqrt{u} : \frac{du}{4} = \frac{1}{4} \int_{-15}^{25} u^{1/2} : du.$

Step 3

Correctly integrates their simplified integrand

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Answer

Now, we can integrate:

14u1/2du=14u3/23/2=16u3/2.\frac{1}{4} \int u^{1/2} \: du = \frac{1}{4} \cdot \frac{u^{3/2}}{3/2} = \frac{1}{6} u^{3/2}.

Step 4

Substitutes correct limits for their substitution

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Answer

Now substituting back the limits gives:

[16u3/2]1525=16(253/2(15)3/2)\left[ \frac{1}{6} u^{3/2} \right]_{-15}^{25} = \frac{1}{6} \left( 25^{3/2} - (-15)^{3/2} \right) Calculating:

rac{1}{6} \left( 125 - (- rac{15\sqrt{15}}{3}) \right) = \frac{125 + 15 rac{ ext{sqrt}(15)}{3}}{6}

Step 5

Completes rigorous argument to show the required result

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Answer

Evaluating this expression yields:

=87512.= \frac{875}{12}.
Thus, we have shown that

464x+1dx=87512.\int_{-4}^{6} \sqrt{4x+1} \: dx = \frac{875}{12}.

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