Techniques & Applications of Integration (DP IB Analysis & Approaches (AA): HL): Exam Questions

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 $y=f\left(x\right)$ intercepts the x-axis.

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

Find the indefinite integral for

$\int \cos \left(\frac{x}{2}\right)\mathrm{d}x$

Find the indefinite integral for

$\int 5e^{3x}\mathrm{d}x$

Find an expression for y given that

$\frac{\mathrm{d}y}{\mathrm{d}x}=\sin \left(x-\frac{\mathrm{\pi }}{3}\right)$

Find the exact value of

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

Find the definite integral

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

Find an expression for y given that

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

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

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

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

(ii) Hence, find

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

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

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

Use a suitable substitution to show that

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

Using a suitable trigonometric identity, find the exact value of

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

Work out the value of the following definite integral

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

giving your answer as an exact value.

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

The shaded region in the diagram is the area bounded 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.

The diagram below shows a sketch of part of the curves with equations  

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

The shaded region in the diagram is the area bounded by the two curves.

Work out the area of the region bounded by the positive x-axis, the negative y-axis and the graph of  $y=2x^3-21x^2+66x-47$

Work out the area of the shaded region.

Consider the function $h\left(x\right)$such that

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

(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

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

Find

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

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

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

Find the indefinite integral for

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

Find the indefinite integral for

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

Find an expression for y given that

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

Find the exact value of

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

Find the definite integral

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

Find an expression for y given that

$\frac{dy}{dx}=x{e}^{x^2-2}$

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

Use a suitable substitution to show that

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

Let I be the definite integral defined by

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

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

Explain why

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

Use the result from part (a) to show that

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

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

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

The shaded region in the diagram is the area bounded by the two curves.

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

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

calculate 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.

The diagram shows a sketch of 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 diagram below shows a sketch of 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.  Be sure to provide a suitable justification for your answer.

Consider the function $h\left(x\right)$ 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

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

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.

Determine the value of I, giving your answer in terms of a.

Hence, or otherwise, determine the value of a which maximises the value of I, and calculate the value of I when a takes that value.