P2 Integration: The Reverse of Differentiation and Area Calculation

Let's Put the Pieces Back Together! 🧩

Introduction

1. Introduction

Alright, let's dive into one of the most fundamental concepts in A-Level Maths: Integration! Think of it as the ultimate 'undo' button for differentiation. In this chapter, we're going to explore how to reverse the process of finding gradients to instead find the area under a curve. We'll kick things off by integrating standard functions like polynomials, exponentials, and trig functions. Then, we'll level up by using our trusty trigonometric identities to simplify and solve more complex integrals. Finally, for those tricky curves that can't be integrated perfectly, we'll learn a powerful approximation technique called the Trapezium Rule . By the end of this, you'll have a solid toolkit for tackling a huge range of exam questions. Let's do this!

e^{3x+5}

\int e^{ax+b} dx = \frac{1}{a}e^{ax+b} + C

Worked example

Let's Actually Solve One of These Beasts 🐲

I = \int_{0}^{\frac{\pi}{8}} 3\sec^2(2x) dx

1
$\sec^2(ax+b)$
$\text{Standard form: } \int \sec^2(ax+b) dx = \frac{1}{a}\tan(ax+b) + C$
2
$\sec^2(2x)$
$\int 3\sec^2(2x) dx = 3 \times \left( \frac{1}{2}\tan(2x) \right) = \frac{3}{2}\tan(2x)$
3
$Since this is a definite integral, we don't need the '+ C'. We place our result in square brackets and write the limits of integration. This sets us up to substitute the values in.$
$I = \left[ \frac{3}{2}\tan(2x) \right]_{0}^{\frac{\pi}{8}}$
4
$x=\frac{\pi}{8}$
$I = \left( \frac{3}{2}\tan\left(2 \times \frac{\pi}{8}\right) \right) - \left( \frac{3}{2}\tan(2 \times 0) \right)$
5
$Let's simplify the expressions inside the tan functions. This is where knowing your exact trig values comes in clutch for getting that 'exact value' the question demands.$
$I = \left( \frac{3}{2}\tan\left(\frac{\pi}{4}\right) \right) - \left( \frac{3}{2}\tan(0) \right)$
6
$\tan(\frac{\pi}{4})=1$
$I = \left( \frac{3}{2} \times 1 \right) - \left( \frac{3}{2} \times 0 \right) = \frac{3}{2}$

Answer

I = \left( \frac{3}{2} \times 1 \right) - \left( \frac{3}{2} \times 0 \right) = \frac{3}{2}

y = e^{-x^2}

Now for the crucial exam detail: overestimate vs. underestimate . If the curve is concave up (it curves upwards, like a 'U'), the straight tops of the trapeziums will sit above the curve, giving you an overestimate . If the curve is concave down (it curves downwards, like an 'n'), the trapezium tops will sit below the curve, resulting in an underestimate . Knowing this is key for those tricky 'comment on your answer' questions. This concept is foundational for university-level engineering, physics, and data science, where you'll constantly be approximating solutions to complex problems.

Worked example

Let's Calculate Some Area! 🚀

\int_{1}^{3} \sqrt{1+x^3} \, dx

1
$h$
$h = \frac{3 - 1}{4} = \frac{2}{4} = 0.5$
2
$x_0=1$
$x_0 = 1, \quad x_1 = 1.5, \quad x_2 = 2, \quad x_3 = 2.5, \quad x_4 = 3$
3
$y = \sqrt{1+x^3}$
$y_0 = \sqrt{1+1^3} = \sqrt{2} \approx 1.41421 y_1 = \sqrt{1+1.5^3} = \sqrt{4.375} \approx 2.09165 y_2 = \sqrt{1+2^3} = \sqrt{9} = 3 y_3 = \sqrt{1+2.5^3} = \sqrt{16.625} \approx 4.07738 y_4 = \sqrt{1+3^3} = \sqrt{28} \approx 5.29150$
4
$Time to substitute everything into the Trapezium Rule formula. Remember, it's the first y-value, plus the last y-value, plus twice the sum of all the middle ones.$
$\text{Area} \approx \frac{1}{2}(0.5) [ (1.41421 + 5.29150) + 2(2.09165 + 3 + 4.07738) ]$
5
$Calculate the final result and round to the required 3 significant figures. Don't forget this last step - easy marks to lose!$
$\text{Area} \approx 0.25 [ 6.70571 + 2(9.16903) ] \approx 0.25 [ 6.70571 + 18.33806 ] \approx 0.25 [ 25.04377 ] \approx 6.26094 \approx 6.26 \quad (3 \text{ s.f.})$
6
$y = \sqrt{1+x^3}$
$y = \sqrt{1+x^3}$

Answer

y = \sqrt{1+x^3}

\sin^2 x

\sin^2 x = \frac{1}{2}(1 - \cos(2x))

Worked example

Let's Solve This Thing! 🤓

\int_{0}^{\frac{\pi}{2}} \sin^2 x \, dx

1
$\sin^2 x$
$\text{Use the identity: } \cos(2x) = 1 - 2\sin^2 x$
2
$\sin^2 x$
$2\sin^2 x = 1 - \cos(2x) \implies \sin^2 x = \frac{1}{2}(1 - \cos(2x))$
3
$\sin^2 x$
$\int_{0}^{\frac{\pi}{2}} \sin^2 x \, dx = \int_{0}^{\frac{\pi}{2}} \frac{1}{2}(1 - \cos(2x)) \, dx$
4
$\frac{1}{2}$
$\frac{1}{2} \left[ x - \frac{1}{2}\sin(2x) \right]_{0}^{\frac{\pi}{2}}$
5
$\frac{\pi}{2}$
$\frac{1}{2} \left[ \left( \frac{\pi}{2} - \frac{1}{2}\sin(\pi) \right) - \left( 0 - \frac{1}{2}\sin(0) \right) \right]$
6
$\sin(\pi) = 0$
$\frac{1}{2} \left[ \left( \frac{\pi}{2} - 0 \right) - (0 - 0) \right] = \frac{1}{2} \times \frac{\pi}{2} = \frac{\pi}{4}$

Answer

\frac{1}{2} \left[ \left( \frac{\pi}{2} - 0 \right) - (0 - 0) \right] = \frac{1}{2} \times \frac{\pi}{2} = \frac{\pi}{4}

Unlock the full chapter: practice questions, flashcards, mock papers and notes, free.

Continue revising