Mathematics is the easiest subject to increase the score if practiced well. Grab this model question paper for 11th state board Maths and practice to get the high score in exams. In this question paper, questions are prepared based on the state board syllabus with the book back and creative questions. This is the most important model question paper for plus one maths for the chapter Integral Calculus.

Creative questions were added in this question paper in order to enhance the student's knowledge and get the idea that how to answer strategic questions asked in the board exams. Now you can use this model question paper for practice on your own to score more on your board exams. By practicing with this important model question paper student can gain more knowledge and confidence to crack the exams. Our aim is to bring out the best scores on board exams. This question paper is designed based on the academic syllabus. It will be more helpful for the Matriculation and state board students to prepare for their exams.

This question paper exclusively designed for the chapter Integral Calculus from plus one Maths.HSC Board Exams are fast approaching and students are getting anxious about how to prepare for their HSC Board Exams. The latest sample papers have been designed as per the latest syllabus and examination trends. Students can make use of this question paper since all the questions are taken from the book back and creative question papers.

Download Tamil Nadu 11th Standard Maths question papers, model tests, one-mark questions, important questions, and public exam papers in PDF format. Free study materials and answer keys for TN State Board students.

Questions:

Generic 2

6 x 2 = 12

Evaluate : \(\int { { x }^{ 2 } } \) log xdx

Evaluate : \(\int { \sqrt { x } -{ cos }^{ 2 } } \frac { x }{ 2 } \)

Integrate the following with respect to x : \(\left(1-x^2\right)^{-\frac{1}{2}}\)

Integrate the following with respect to x : \({x^{24}\over x^{25}}\)

Integrate the following with respect to x : \({1\over sin ^2 x}\)

Integrate the following with respect to x : \(\sqrt{x}\)

Generic 3

5 x 3 = 15

Evaluate \(\int { \frac { 1 }{ { 4x }^{ 2 }-4x+3 } } \)dx

Evaluate \(\int { \frac { { sin }^{ 6 }x+cos^{ 6 }x }{ sin^{ 2 }xcos^{ 2 }x } } \)

Integrate the following with respect to x : 9xe^3x

10.

Evaluate : \(\int (x-3)\sqrt{x+2}dx.\)

11.

Integrate the following functions with respect to x : \({1\over 6\ -\ 4x}\)

Multiple Choice Question

10 x 1 = 10

12.

\(\int { x } \) sin x dx = -x cos x + a, then a =

(a)

sin x + c

(b)

cos x + c

(c)

(d)

none of these

13.

\(\int { \frac { { e }^{ x } }{ \left( 1+{ e }^{ x } \right) ^{ 2 } } } \) dx =_______+c

(a)

\(\frac { 1 }{ 1+{ e }^{ x } } \)

(b)

-\(\frac { 1 }{ 1+{ e }^{ x } } \)

(c)

1 + e^x

(d)

\(\frac { \left( 1+{ e }^{ x } \right) ^{ 3 } }{ 3 } \)

14.

\(\int { \frac { \left( 1+logx \right) ^{ 2 } }{ x } } \) dx = _______+C.

(a)

\(\frac { \left( 1+logx \right) ^{ 3 } }{ 3 } \)

(b)

3 log ( 1 + log x)

(c)

2(1 + log x)

(d)

none of these

15.

\(\int { { 3 }^{ x+2 } } \) dx = __________+c.

(a)

\(\frac { { 3 }^{ x } }{ log3 } \)

(b)

\(9\left( \frac { { 3 }^{ x } }{ log3 } \right) \)

(c)

\(\frac { { 3 }^{ x } }{ 9log3 } \)

(d)

3^x.9

16.

\(\int { \frac { 1 }{ 2 } } { sec }^{ 2 }\) x dx is _____ + c.

(a)

\(\frac { 1 }{ 2 } \) tan x

(b)

tan x

(c)

2 tan x

(d)

none of these

17.

\(\int \frac{1}{x \sqrt{(\log x)^2-5}} d x\) is

(a)

\(log|x+\sqrt{x^2-5}|+c\)

(b)

\(log|logx+\sqrt{logx-5}|+c\)

(c)

\(log|logx+\sqrt{(logx)^2-5}|+c\)

(d)

\(log|logx-\sqrt{(logx)^2-5}|+c\)

18.

\(\int \frac{\sec ^2 x}{\tan ^2 x-1} d x\) is

(a)

\(2log|{1-tan x\over 1+tan \ x}|+c\)

(b)

\(log|{1+tan x\over 1-tan \ x}|+c\)

(c)

\({1\over2}log|{tan x+1\over tan \ x-1}|+c\)

(d)

\({1\over2}log|{tan x-1\over tan \ x+1}|+c\)

19.

\(\int \frac{d x}{e^x-1}\) is

(a)

\(log|e^x|-log|e^x-1|+c\)

(b)

\(log|e^x|+log|e^x-1|+c\)

(c)

\(log|e^x-1|-log|e^x|+c\)

(d)

\(log|e^x+1|-log|e^x|+c\)

20.

\(\int \sin ^3 x d x\) is

(a)

\({-3\over 4}cos \ x-{cos \ 3x\over 12}+c\)

(b)

\({3\over 4}cos \ x+{cos \ 3x\over 12}+c\)

(c)

\({-3\over 4}cos \ x+{cos \ 3x\over 12}+c\)

(d)

\({-3\over 4}sin \ x-{sin \ 3x\over 12}+c\)

21.

If \(\int {3^{1\over x}\over x^2}dx=k(3^{1\over x})+c\), then the value of k is

(a)

log 3

(b)

-log 3

(c)

\(-{1\over log3}\)

(d)

\({1\over log3}\)

Generic 5

2 x 5 = 10

22.

Evaluate \(\int { \frac { { e }^{ x }dx }{ { e }^{ 2x }+6{ e }^{ x }+5 } } \)

23.

Evaluate \(\int { \frac { dx }{ tanx+cotx+secx+cosecx } } \)

Visit QB365-SLMS - https://www.qb365.in/slms/

Answers Key:

Generic 2

6 x 2 = 12

Let I = \(\int { { x }^{ 2 } } \) log x dx
u = log x; dv = x² dx
du = \(\frac { 1 }{ x } \)dx; V = \(\frac { { x }^{ 3 } }{ 3 } \)
\(\therefore\) Applying integration by parts , we get
\(\int { u } dv=uv-\int { v } du\)
\(\therefore\) I = \(\int { { x }^{ 2 } } \) log x dx = \(\frac { { x }^{ 3 } }{ 3 } \)log x - \(\int { \frac { { x }^{ 3 } }{ 3 } } .\frac { 1 }{ 3 } \)dx \(\frac { { x }^{ 3 } }{ 3 } \)log x - \(\frac { 1 }{ 3 } \)\(\int { { x }^{ 2 } } \)dx
I = \(\frac { { x }^{ 3 } }{ 3 } \)log x - \(\frac { 1 }{ 3 } \)\(\left( \frac { { x }^{ 3 } }{ 3 } \right) \) + 1
= \(\frac { { x }^{ 3 } }{ 3 } \)log x - \(\frac { { x }^{ 3 } }{ 9 } \) (3log x - 1) +c

\(\int { \left( \sqrt { x } -{ cos }^{ 2 }\frac { x }{ 2 } \right) } \)dx = \(\int { { x }^{ \frac { 1 }{ 2 } }dx- } \int { { cos }^{ 2 }\frac { x }{ 2 } dx } \)
= \(\frac { { x }^{ \frac { 1 }{ 2 } } }{ \frac { 1 }{ 2 } +1 } -\int { \frac { 1+cosx }{ 2 } dx= } \)\(\frac { { x }^{ \frac { 3 }{ 3 } } }{ \frac { 3 }{ 2 } } -\int { \frac { 1 }{ 2 } dx } -\int { \frac { cosx }{ 2 } } dx\)
=\(\frac { 2 }{ 3 } { x }^{ \frac { 3 }{ 2 } }-\frac { 1 }{ 2 } x-\frac { 1 }{ 2 } \)sin x + c
= \(\frac { 2 }{ 3 } { x }^{ \frac { 3 }{ 2 } }-\frac { 1 }{ 2 } \) (x +sin x) + c

\( \int\left(1-x^2\right)^{-\frac{1}{2}} d x =\int \frac{1}{\left(1-x^2\right)^{1 / 2}} d x \)
\(=\int \frac{1}{\sqrt{1-x^2}} d x=\sin ^{-1} x+c\)

\(\int { \frac { { x }^{ 24 } }{ { x }^{ 25 } } } dx=\int { \frac { 1 }{ x } } dx\)
= \(log|x|+c\)

\(\int { \frac { 1 }{ { sin }^{ 2 }x } } dx=\int { cosec^{ 2 } } xdx\)
= − cot x +c

\(\int {\sqrt{x}\ dx }= \int { {x^{1\over 2}} }dx={x^{{1\over 2}+1}\over {1\over2}+1}+c={x^{3\over 2}\over {3\over2}}+c={2\over3}x^{3\over2}+c\)

Generic 3

5 x 3 = 15

Let I = \(\int { \frac { 1 }{ { 4x }^{ 2 }-4x+3 } } \)dx = \(\int { \frac { dx }{ \left( { x }^{ 2 }-x+\frac { 3 }{ 4 } \right) } } \)dx
= \(\frac { 1 }{ 4 } \int { \frac { dx }{ { x }^{ 2 }-x+\frac { 1 }{ 4 } -\frac { 1 }{ 4 } +\frac { 3 }{ 4 } } } =\frac { 1 }{ 4 } \int { \frac { 1 }{ \left( x-\frac { 1 }{ 2 } \right) ^{ 2 }+\left( \frac { 1 }{ \sqrt { 2 } } \right) ^{ 2 } } } \)dx
I = \(\frac { 1 }{ 4 } \times \frac { 1 }{ \frac { 1 }{ \sqrt { 2 } } } { tan }^{ -1 }\left( \frac { x-\frac { 1 }{ \sqrt { 2 } } }{ \frac { 1 }{ \sqrt { 2 } } } \right) +c\)
\(\therefore\) I = \(\frac { 1 }{ 2\sqrt { 2 } } { ta }n^{ -1 }\left( \frac { 2x-1 }{ \sqrt { 2 } } \right) +c\)

= \(\int { \frac { { sin }^{ 6 }x+cos^{ 6 }x }{ sin^{ 2 }xcos^{ 2 }x } } \)
= \(\int { \frac { \left( { sin }^{ 2 }x \right) ^{ 3 }+\left( { cos }^{ 2 }x \right) ^{ 3 } }{ { sin }^{ 2 }x{ cos }^{ 2 }x } } \) dx\(\left[ { a }^{ 3 }+{ b }^{ 3 }=\left( a+b \right) ^{ 3 }-3ab\left( a+b \right) \right] \)
= \(\int { \frac { \left( { sin }^{ 2 }x \right) ^{ 3 }+\left( { cos }^{ 2 }x \right) ^{ 3 }-{ 3sin }^{ 2 }x{ cos }^{ 2 }x\left( { sin }^{ 2 }x+{ cos }^{ 2 }x \right) }{ { sin }^{ 2 }x{ cos }^{ 2 }x } } \)
= \(\int { \frac { { 1 }^{ 3 }-3{ sin }^{ 2 }x{ cos }^{ 2 }x(1) }{ { sin }^{ 2 }x{ cos }^{ 2 }x } } \) dx = \(\int { \frac { 1-3{ sin }^{ 2 }x{ cos }^{ 2 }x }{ { sin }^{ 2 }x{ cos }^{ 2 }x } } \)dx
= \(\int { \left( \frac { 1 }{ { sin }^{ 2 }x{ cos }^{ 2 }x } -3 \right) } \)dx = \(\int { \frac { { sin }^{ 2 }x{ +cos }^{ 2 }x }{ { sin }^{ 2 }x{ cos }^{ 2 }x } } -3\)dx
= \(\int { \frac { { sin }^{ 2 }x }{ { sin }^{ 2 }x{ cos }^{ 2 }x } } dx+\int { \frac { { cos }^{ 2 }x }{ { sin }^{ 2 }x{ cos }^{ 2 }x } } dx-\int { 3 } \)dx
= \(\int { sec^{ 2 } } \) x dx + \(\int { cosec } ^{ 2 }xdx\) - 3 \(\int { dx } \) = tan x - cot x - 3x + c

\(\int 9 x e^{3 x} d x=9 \int x e^{3 x} d x \)
Here, \(u=x\ and\ d v=e^{3 x} \)
\(\Rightarrow d u=d xand v=\frac{e^{3 x}}{3}\)
Applying integration by parts, we get
\( =9\left[x\left(\frac{e^{3 x}}{3}\right)-\int \frac{e^{3 x}}{3} \times d x\right] \)
\(=9\left[x\left(\frac{e^{3 x}}{3}\right)+\frac{1}{3}\left(\frac{e^{3 x}}{3}\right)\right]+c \)
\(=3 x e^{3 x}-e^{3 x}+c \)
\(=(3 x-1) e^{3 x}+c \)
Aliter: Using Bernoullis formula
\( \int 9 x e^{3 x} d x =9 x\left(\frac{e^{3 x}}{3}\right)-9\left(\frac{e^{3 x}}{3 \times 3}\right)+c \)
\(=3 x e^{3 x}-e^{3 x}+c \)
\(=(3 x-1) e^{3 x}+c \)

10.

\(\int (x-3)\sqrt{x+2}dx=\int (x+2-5)\sqrt{x+2}dx\)
\(=\int (x+2)\sqrt{x+2}dx-5\int\sqrt{x+2}dx\)
\(=\int(x+2)^{3\over2}dx-5\int (x+2)^{1\over2}dx\)
\(={(x+2)^{5\over2}\over {5\over2}}-5{(x+2)^{3\over2}\over {3\over2}}+c\)
\(={2\over5}(x+2)^{5\over2}-{10\over 3}(x+2)^{3\over2}+c\)

11.

\(\int \frac{1}{x} d x=\log |x|+c\)
\(\int { \frac { 1 }{ 6-4x } } dx=\frac { log|6-4x| }{ -4 } +c\)
= \(-{1\over4}log|6-4x|+c\)

Multiple Choice Question

10 x 1 = 10

12.

(a)
sin x + c

13.

(b)
-\(\frac { 1 }{ 1+{ e }^{ x } } \)

14.

(a)
\(\frac { \left( 1+logx \right) ^{ 3 } }{ 3 } \)

15.

16.

(a)
\(\frac { 1 }{ 2 } \) tan x

17.

\(\int \frac{1}{x \sqrt{(\log x)^{2}-5}} d x =\int \frac{1}{\sqrt{(\log x)^{2}-5}} \times \frac{1}{x} d x \)
\(=\int \frac{1}{\sqrt{u^{2}-5}} d u, \quad u=\log x \)
\(=\log \left|u+\sqrt{u^{2}-5}\right|+c \)
\(=\log \left|\log x+\sqrt{(\log x)^{2}-5}\right|+c \)

18.

\(\int \frac{\sec ^{2} x}{\tan ^{2} x-1} d x =\int \frac{1}{(\tan x)^{2}-1} \times \sec ^{2} x d x \)
\(=\int \frac{1}{u^{2}-1} d u, \quad u=\tan x \)
\(=\frac{1}{2(1)} \log \left|\frac{u-1}{u+1}\right|+c \)
\(\frac{1}{2} \log \left|\frac{\tan x-1}{\tan x+1}\right|+c \)

19.

\(\int \frac{d x}{e^{x}-1} =\int \frac{1}{e^{x}\left(1-\frac{1}{e^{x}}\right)} d x \)
\(=\int \frac{e^{-x}}{1-e^{-x}} d x \)
\(=\log \left|1-e^{-x}\right|+c \)
\(=\log \left|1-\frac{1}{e^{x}}\right|+c \)
\(=\log \left|\frac{e^{x}-1}{e^{x}}\right|+c \)
\(=\log \left|e^{x}-1\right|-\log \left|e^{x}\right|+c \)

20.

\(\sin 3 x =3 \sin x-4 \sin ^{3} x \)
\(\therefore 4 \sin ^{3} x =3 \sin x-\sin 3 x \)
\(\therefore \sin ^{3} x =\frac{1}{4}[3 \sin x-\sin 3 x] \)
\(\therefore \int \sin ^{3} x d x =\frac{1}{4}\left[-3 \cos x+\frac{\cos 3 x}{3}\right]+c \)
\(=\frac{-3}{4} \cos x+\frac{\cos 3 x}{12}+c \)

21.

Generic 5

2 x 5 = 10

22.

Let I = \(\int { \frac { { e }^{ x }dx }{ { e }^{ 2x }+6{ e }^{ x }+5 } } \) = \(\int { \frac { { e }^{ x }dx }{ \left( { e }^{ x } \right) ^{ 2 }+6{ e }^{ x }+5 } } \)
Put t= e^x \(\Rightarrow\) dt = e^x dx
I = \(\int { \frac { dt }{ { t }^{ 2 }+6t+5 } =\int { \frac { dt }{ { t }^{ 2 }+6t+9-9+5 } } =\int { \frac { dt }{ \left( t+3 \right) ^{ 2 }-4 } } \int { \frac { dt }{ \left( t+3 \right) ^{ 2 }-{ 2 }^{ 2 } } } } \)
= \(\frac { 1 }{ 2\left( 2 \right) } log\left| \frac { t+3-2 }{ t+3+2 } \right| +c\)
= \(\frac { 1 }{ 4 } log\left| \frac { { e }^{ x }+1 }{ { e }^{ x }+5 } \right| +c\)

23.

Let I = \(\int { \frac { dx }{ tanx+cotx+secx+cosecx } } \)
= \(\int { \frac { dx }{ \frac { sinx }{ cosx } +\frac { cosx }{ sinx } +\frac { 1 }{ sinx } +\frac { 1 }{ cosx } } } \) = \(\int { \frac { sincosxdx }{ { sin }^{ 2 }x+{ cos }^{ 2 }+cosx+sinx } } \)
= \(\frac { 1 }{ 2 } \int { \frac { 2sinxcosdx }{ 1+sinx+cosx } } \) = \(\frac { 1 }{ 2 } \int { \frac { (1+2sinxcos)-1 }{ 1+sinx+cosx } } \) dx
= \(\frac { 1 }{ 2 } \int { \frac { \left( { sin }^{ 2 }x+{ cos }^{ 2 }x+2sinxcosx \right) -1 }{ 1+sinx+cosx } } \) dx = \(\frac { 1 }{ 2 } \int { \frac { \left( sinx+cosx \right) ^{ 2 }-{ 1 }^{ 2 } }{ 1+sinx+cosx } } \)dx
= \(\frac { 1 }{ 2 } \int { \left( sinx+cox-1 \right) } \)
= \(\frac { 1 }{ 2 } \) (-cos x + sin x - x) + c