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###### A-Level H2 Math | 5 Essential Questions

# Applications of Integration

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**2013 RVHS Promo Q3**

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**2013 AJC Promo Q8 (b)**

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**2017 YJC P1 Q9**

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**2020 ASRJC P2 Q4**

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**2011 CJC P1 Q8**

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Integration is a very useful extension of differentiation and relatively easy to understand. They are also known as the opposite of the derivatives. The definite integral is a powerful concept used to represent the area under a curve, which turns out to be useful in a lot of calculations.Â

- Q1
- Q2
- Q3
- Q4
- Q5

The diagram shows the sketch of the curve $C$, ${{\left( y-1 \right)}^{2}}=x\sqrt{{{x}^{2}}-1}$, with the vertex at $\left( 1,1 \right)$.

(i)

Write down the equation of the graph when $C$ is translated $1$ unit in the negative $y$-direction.

[1]

(i) Write down the equation of the graph when $C$ is translated $1$ unit in the negative $y$-direction.

[1]

(ii)

The shaded region $R$, bounded by $C$ and the vertical line, $x=a$, is rotated through $\pi $ radians about the line $y=1$. By using the substitution $u=\sqrt{{{x}^{2}}-1}$, or otherwise, find the exact volume obtained in terms of $a$.

[5]

(ii) The shaded region $R$, bounded by $C$ and the vertical line, $x=a$, is rotated through $\pi $ radians about the line $y=1$. By using the substitution $u=\sqrt{{{x}^{2}}-1}$, or otherwise, find the exact volume obtained in terms of $a$.

[5]

- (i)
- (ii)

- (i)
- (ii)

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The diagram shows a shaded region $R$ bounded by the curve ${{\left( y-2 \right)}^{2}}=x+1$ and the line $y+2x=6$.

Find the volume generated when $R$ is rotated through $2\pi $ radians about the $x$-axis, leaving your answer correct to $3$ significant figures.[4]

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(a)

By using the substitution $x=3\sec \theta $, evaluate $\int_{3\sqrt{2}}^{6}{\frac{3x+1}{\sqrt{{{x}^{2}}-9}}\text{d}x}$ exactly.

[5]

(a) By using the substitution $x=3\sec \theta $, evaluate $\int_{3\sqrt{2}}^{6}{\frac{3x+1}{\sqrt{{{x}^{2}}-9}}\text{d}x}$ exactly.

[5]

(b)

The diagram shows an ellipse with equation $\frac{{{x}^{2}}}{16}+\frac{{{\left( y-2 \right)}^{2}}}{4}=1$.

(b)

The diagram shows an ellipse with equation $\frac{{{x}^{2}}}{16}+\frac{{{\left( y-2 \right)}^{2}}}{4}=1$.

(i)

Find the area of the shaded region, giving your answer correct to $3$ decimal places.

[2]

(i) Find the area of the shaded region, giving your answer correct to $3$ decimal places.

[2]

(ii)

Find the exact volume of the solid generated when the shaded region is rotated $180{}^\circ $ about the $y$-axis.

[4]

(ii) Find the exact volume of the solid generated when the shaded region is rotated $180{}^\circ $ about the $y$-axis.

[4]

- (a)
- (b) (i)
- (b) (ii)

- (a)
- (b) (i)
- (b) (ii)

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(a)

Find $\int{\left( {{\cot }^{6}}2x+{{\cot }^{4}}2x-\sin \frac{x}{2}\sin \frac{3x}{2} \right)}\text{ d}x$.

[3]

(a) Find $\int{\left( {{\cot }^{6}}2x+{{\cot }^{4}}2x-\sin \frac{x}{2}\sin \frac{3x}{2} \right)}\text{ d}x$.

[3]

(b)

The diagram shows the graphs of $y={{\text{e}}^{2x}}$ and $y={{\left( x+1 \right)}^{2}}$.
$R$ is the finite region bounded by the two curves $y={{\text{e}}^{2x}}$ and $y={{\left( x+1 \right)}^{2}}$.
Find the volume of the solid formed when $R$ is rotated through $2\pi $ radians about the $y$-axis, giving your answer correct to $4$ decimal places.

[3]

(b) The diagram shows the graphs of $y={{\text{e}}^{2x}}$ and $y={{\left( x+1 \right)}^{2}}$.

$R$ is the finite region bounded by the two curves $y={{\text{e}}^{2x}}$ and $y={{\left( x+1 \right)}^{2}}$.

Find the volume of the solid formed when $R$ is rotated through $2\pi $ radians about the $y$-axis, giving your answer correct to $4$ decimal places.

[3]

(c)

A curve $C$ has parametric equations

$x=2\left( \sin \frac{t}{2}-\cos \frac{t}{2} \right)$, $y=\sin \frac{t}{2}+\cos \frac{t}{2}$, for $-\frac{3\pi }{2}\le t\le -\frac{\pi }{2}$.

The line $L$ with equation $y=-2x-5$ meets the curve $C$ at the point $\left( -2,\,-1 \right)$. $S$ is the region enclosed by line $L$, curve $C$and the $x$-axis as shown in the diagram below.(c) A curve $C$ has parametric equations

$x=2\left( \sin \frac{t}{2}-\cos \frac{t}{2} \right)$, $y=\sin \frac{t}{2}+\cos \frac{t}{2}$, for $-\frac{3\pi }{2}\le t\le -\frac{\pi }{2}$.

The line $L$ with equation $y=-2x-5$ meets the curve $C$ at the point $\left( -2,\,-1 \right)$.

$S$ is the region enclosed by line $L$, curve $C$and the $x$-axis as shown in the diagram below.

(i)

Show that the x-intercept of curve $C$ is $-2\sqrt{2}$.

[2]

(i) Show that the x-intercept of curve $C$ is $-2\sqrt{2}$.

[2]

(ii)

Find the exact area of $S$.

[6]

(ii) Find the exact area of $S$.

[6]

- (a)
- (b)
- (c) (i)
- (c) (ii)

- (a)
- (b)
- (c) (i)
- (c) (ii)

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The diagram below shows the curve with equation $y={{x}^{2}}-1$, $x>0$ and the tangent $y=2x-2$ to the curve at the point $(1,0)$.

(i)

Write down the coordinates of the point where this tangent meets the line $x=3$.

[1]

(i) Write down the coordinates of the point where this tangent meets the line $x=3$.

[1]

The region, $R$, shaded in the diagram, is bounded by the curve, the tangent to the curve at $(1,0)$ and the line $x=3$.

Find

(ii)

the area of $R$,

[2]

(ii) the area of $R$,

[2]

(iii)

the exact volume of the solid of revolution obtained when $R$ is rotated through $2\pi $ radians about the $y$-axis.

[4]

(iii) the exact volume of the solid of revolution obtained when $R$ is rotated through $2\pi $ radians about the $y$-axis.

[4]

- (i)
- (ii)
- (iii)

- (i)
- (ii)
- (iii)

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A curve $C$ has equation $y=\frac{1}{\sqrt{4-{{x}^{2}}}}$ for $-1\le x\le 1$. The region $R$ is enclosed by $C$, the $x$- axis and the lines $x=-1$ and $x=1$.

(i)

Find the exact value of the area of $R$.

[3]

(ii)

Find the exact value of the volume generated when $R$ is rotated through two right angles about the $x$- axis.

[3]

(iii)

Show that the volume generated when $R$ is rotated through two right angles about the $y$- axis is $\pi \,(4-2\sqrt{3})$.

[4]

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