Further Integration (DP IB Analysis & Approaches (AA): SL): Exam Questions

Integrate

$\int \cos 2x\mathrm{d}x$

Find the definite integral

${\int }_0^2{\left(3x-1\right)}^3\mathrm{d}x$

Find an expression for $y$ given that

$\frac{\mathrm{d}y}{\mathrm{d}x}={\mathrm{e}}^{5x}$

Show that $\frac{x}{x+4}$ can be written as $1-\frac{4}{x+4}$.

Hence show that

${\int }_1^2\frac{x}{x+4}\mathrm{d}x=1+4\ln \frac{5}{6}$

Given that

$\cos 2\theta \equiv 2{\cos }^2\theta -1$

use calculus to find the exact value of

${\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}{\cos }^2\theta \mathrm{d}\theta$

Given that $f\left(x\right)=2x^3+4x$, find $f\text{'}\left(x\right)$.

Hence, or otherwise, find

$\int \frac{3x^2+2}{2x^3+4x}\mathrm{d}x$

The following diagram shows the curves with equations

$y=x^2-3x+4$ and $y=4-x^2+2x$

Find the $x$-coordinates of the points of intersection of the two curves.

Show that the area of the shaded region $R$ is given by

${\int }_0^{\frac{5}{2}}\left(5x-2x^2\right)\mathrm{d}x$

Use calculus to find the area of $R$.

The following diagram shows the graphs of the line $y=6-x$ and the curve $y=x^2$.

Point P is the point of intersection of the curve $y=x^2$ with the $x$-axis. Point Q is the point of intersection of the curve $y=x^2$ with the line $y=6-x$ for which $x>0$. Point R is the point of intersection of the line $y=6-x$ with the $x$-axis.

Find the $x$-coordinates of P, Q and R.

Find the area of the shaded region.

Consider the function $h$ such that

${\int }_1^{5}h\left(x\right)\mathrm{d}x=2$

Find

${\int }_5^{1}h\left(x\right)\mathrm{d}x$

Find

${\int }_1^5\frac{h\left(x\right)+1}{2}\mathrm{d}x$

Find

${\int }_1^5\left(h\left(x\right)+2x\right)\mathrm{d}x$

Consider the function $f\left(x\right)=\ln \left(2x^2+1\right)$.

Find $f\text{'}\left(x\right)$.

Hence, find

$\int \frac{x}{2x^2+1}\mathrm{d}x$

Let $f\text{'}\left(x\right)=x^2\cos \left(x^3+1\right)$.

Find $f\left(x\right)$, given that $f\left(-1\right)=1$.

Given that

${\int }_k^5\left(2x-1\right)\mathrm{d}x=20,k>0$

find the value of $k$.

Consider the function $f$ defined by $f\left(x\right)=\left(x^2-x-2\right)\left(x-5\right)$, $-2\le x\le 4$.

Find the coordinates of the points where the graph of $y=f\left(x\right)$ intersects the $x$-axis.

Hence find the area of the region enclosed by the graph of $y=f\left(x\right)$ and the $x$-axis.

Find the exact value of

${\int }_1^5\frac{3}{2x}\mathrm{d}x$

Find the definite integral

${\int }_0^{\frac{\pi }{8}}3\sin 4x\mathrm{d}x$

Find an expression for $y$ given that

$\frac{\mathrm{d}y}{\mathrm{d}x}={\mathrm{e}}^{2x+3}+2$

and also that $y=5$ when $x=-\frac{3}{2}$.

Show that $\frac{x}{2x-3}$ can be written as $\frac{1}{2}+\frac{3}{2\left(2x-3\right)}$.

Hence show that

${\int }_2^5\frac{x}{2x-3}\mathrm{d}x=\frac{3}{2}+\frac{3}{4}\ln 7$

The following diagram shows part of the curves with equations $y=x^2+8x-1$ and $y=4x^2-5x+3$.

The shaded region is enclosed by the two curves.

Show that the area of the shaded region is given by

${\int }_{\frac{1}{3}}^4\left(13x-3x^2-4\right)\mathrm{d}x$

Hence find the area of the shaded region.

Consider the function $h$ such that

${\int }_0^{7}h\left(x\right)\mathrm{d}x=19$ and ${\int }_4^{7}h\left(x\right)\mathrm{d}x=12$

Find

(i) ${\int }_0^{4}h\left(x\right)\mathrm{d}x$

(ii) ${\int }_7^{4}h\left(x\right)\mathrm{d}x$

(iii) ${\int }_3^{3}h\left(x\right)\mathrm{d}x$

Find

${\int }_4^7\frac{4-h\left(x\right)}{5}\mathrm{d}x$

Find

${\int }_0^7\left(2h\left(x\right)+\frac{3x^2}{7}\right)\mathrm{d}x$

Find the indefinite integral for

$\int \sin \left(\frac{\sqrt{3}}{2}x\right)\mathrm{d}x$

Find the indefinite integral for

$\int \frac{7}{{\mathrm{e}}^{4x-9}}\mathrm{d}x$

Find an expression for $y$ given that

$\frac{\mathrm{d}y}{\mathrm{d}x}=\cos \left(2\left(\frac{\pi }{8}-x\right)\right)$

Consider the function $f\left(x\right)=\ln \left(3x^2-12x+1\right)$.

(i) Find $f\text{'}\left(x\right)$.

(ii) Hence, find

$\int \frac{16-8x}{3x^2-12x+1}\mathrm{d}x$

Let $g\text{'}\left(x\right)=\left(x^2-5x+6\right)\sin \left(2x^3-15x^2+36x-\frac{\pi }{3}\right)$.

Find $g\left(x\right)$ given that $g\left(0\right)=1$.

Using a suitable trigonometric identity, find the exact value of

${\int }_{\pi }^{3\pi }{\sin }^2\left(\frac{\theta }{3}\right)\mathrm{d}\theta$

Find the value of

${\int }_2^5\frac{x+1}{x^2+2x-5}\mathrm{d}x$

giving your answer in the form $a\ln b$, where $a,b\in \mathrm{\mathbb{Q}}$.

The following diagram shows part of the curves with equations

$y=2x^3-21x^2+66x-47$ and $y=-3x^3+26x^2-65x+58$

Find the area of the region enclosed by the graph of $y=2x^3-21x^2+66x-47$, the $x$-axis and the $y$-axis.

The shaded region is enclosed by the two curves.

Find the area of the shaded region.

The shaded region $R$ in the following diagram is bounded by the two curves $y=5x^2-12x+8$ and $y=-3x^2+10x+29$.

The two curves intersect at points $\mathrm{A}$ and $\mathrm{B}$ as shown. $x_{\mathrm{A}}$ and $x_{\mathrm{B}}$ are the $x$-coordinates of points $\mathrm{A}$ and $\mathrm{B}$ respectively.

By setting up and solving an appropriate quadratic equation, find the values of $x_{\mathrm{A}}$ and $x_{\mathrm{B}}$.

Find the area of region $R$, giving your answer as an exact value.

Consider the function $f$ defined by $f\left(x\right)=\left(x^2-3x+2\right)\left(x+2\right)$, $x\in \mathrm{ℝ}$.

Find the total area of the regions enclosed by the graph of $y=f\left(x\right)$ and the $x$-axis.

Find the exact value of

${\int }_{-4}^{-1}-\frac{7}{5x}\mathrm{d}x$

Find the definite integral

${\int }_{-\frac{\pi }{3}}^0\sin \left(\frac{\pi }{3}-2x\right)\mathrm{d}x$

Find an expression for $y$ given that

$\frac{\mathrm{d}y}{\mathrm{d}x}=x{\mathrm{e}}^{x^2-2}$

and also that $y=3$ when $x=-\sqrt{2}$.

Express $\frac{x^3}{2\left(x+2\right)\left(x-2\right)}$ in the form $ax+\frac{bx}{x^2-4}$, where $a,b\in \mathrm{\mathbb{Q}}$.

Hence show that

${\int }_3^4\frac{x^3}{2\left(x+2\right)\left(x-2\right)}\mathrm{d}x=\frac{7}{4}+\ln \left(\frac{12}{5}\right)$

The following diagram shows part of the curves with equations $y=x^2-6x+\frac{56}{9}$ and $y=-2x^2+14x-\frac{196}{9}$.

The shaded region is enclosed by the two curves.

By first showing that the area of the shaded region is given by

${\int }_2^{\frac{14}{3}}\left(20x-3x^2-28\right)\mathrm{d}x$

find the exact area of the shaded region.

Explain why your answer to part (a) is not affected by the fact that the shaded region is partially above and partially below the $x$-axis.

Consider the function $h$ such that

${\int }_6^{-3}h\left(x\right)\mathrm{d}x=14$ and ${\int }_2^{5}h\left(x\right)\mathrm{d}x=14$

Find

(i) ${\int }_5^{2}h\left(x\right)\mathrm{d}x$

(ii) ${\int }_{-2}^{-2}h\left(x\right)\mathrm{d}x$

(iii) ${\int }_{-3}^{2}h\left(x\right)\mathrm{d}x+{\int }_5^{6}h\left(x\right)\mathrm{d}x$

Find

${\int }_6^{-3}\frac{7-3h\left(x\right)}{4}\mathrm{d}x$

Given that $h\left(2\right)=3$ and $h\left(5\right)=4$, find

${\int }_2^{5}h\left(x\right)\left(4h\text{'}\left(x\right)-\pi \right)\mathrm{d}x$

Show that $5w^3-21w^2+16=\left(5w+4\right)\left(w^2-5w+4\right)$.

A function $f$ is defined by $f\left(x\right)=-\frac{16}{x^2}-5x+21$, $x\ne 0$.

Let $I$ be the definite integral defined by

$I={\int }_1^{a}f\left(x\right)\mathrm{d}x$

where $a>1$ is a constant.

Find an expression for $I$ in terms of $a$.

Hence, or otherwise, find the value of $a$ which maximises $I$. Justify your answer.

Find this maximum value of $I$.

Let $I$ be the definite integral defined by

$I={\int }_{\frac{a}{k}}^{\frac{b}{k}}{\sin }^2\left(k\theta \right)\mathrm{d}\theta$

where $a$, $b$ and $k$ are real constants such that $a\le b$ and $k>0$.

Show that

$I=\frac{1}{2k}\left[\left(b-a\right)-\frac{1}{2}\left(\sin \left(2b\right)-\sin \left(2a\right)\right)\right]$

Hence find the exact values of

(i) ${\int }_{\frac{\pi }{12}}^{\frac{\pi }{3}}{\sin }^2\left(2\theta \right)\mathrm{d}\theta$

(ii) ${\int }_{-\frac{5\pi }{2}}^{10\pi }{\sin }^2\left(\frac{\theta }{5}\right)\mathrm{d}\theta$

Explain why

$\frac{1}{\tan \theta }=\frac{\cos \theta }{\sin \theta }$

Use the result from part (a) to show that

${\int }_0^{\sqrt{\frac{\pi }{6}}}\frac{x}{\tan \left(x^2-\frac{2\pi }{3}\right)}\mathrm{d}x=-\frac{1}{2}\ln \left(\frac{\sqrt{3}}{2}\right)$

Explain why the value of the integral found in part (b) is a positive number.

The following diagram shows part of the curves with equations $y=2x^3+2x^2-9x-24$ and $y=a{x}^3+b{x}^2+cx+d$, where $a$, $b$, $c$ and $d$ are constants with $a\ne 0$.

The $x$-coordinates of the points of intersection of the two curves are $p$, $q$ and $r$, where $p<q<r$. Region $S$ is the region enclosed by the two curves between $x=p$ and $x=q$, while region $T$ is the region enclosed by the two curves between $x=q$ and $x=r$.

The following diagram shows part of the curve with equation $y=4x^3+9x^2-16x-21$.

The curve intersects the $x$-axis at the points $\left(p,0\right)$, $\left(q,0\right)$ and $\left(r,0\right)$, and region $U$ is the region enclosed by the curve and the $x$-axis between $x=p$ and $x=q$.

Given that the areas of regions $S$ and $U$ are equal, calculate the total area enclosed by the two curves in the first diagram. Justify your answer.