AMU MCA Definite Integration Previous Year Questions (PYQs) – Page 2 of 2
A Place for Latest Exam wise Questions, Videos, Previous Year Papers,
Study Stuff for MCA Examinations
Study Stuff for MCA Examinations
Showing 17 questions
Simpson’s one-third rule for evaluation of $\int_a^b f(x),dx$ requires the interval $[a,b]$ to be divided into:
1
2
3
4
Solution
Simpson’s $\frac{1}{3}$ rule requires an even number of subintervals.
$\int_{1}^{\sqrt{3}} \frac{dx}{1+x^2}$ equals
1
2
3
4
Solution
$ \int \frac{dx}{1+x^2} = \tan^{-1} x $
$ \Rightarrow \int_{1}^{\sqrt{3}} \frac{dx}{1+x^2} = \tan^{-1}(\sqrt{3}) - \tan^{-1}(1) $
$ = \frac{\pi}{3} - \frac{\pi}{4} = \frac{\pi}{12} $
The value of the integral $\int_{0}^{1} \frac{e^{5\log_e x} - e^{4\log_e x}}{e^{\log_e x^3} - e^{\log_e x^2}} , dx$ is
1
2
3
4
Solution
Use $e^{\log_e x^n} = x^n$
So integral becomes:
$\int_{0}^{1} \frac{x^5 - x^4}{x^3 - x^2} , dx$
Factor:
$= \int_{0}^{1} \frac{x^4(x-1)}{x^2(x-1)} , dx$
Cancel $(x-1)$:
$= \int_{0}^{1} x^2 , dx$
$= \left[\frac{x^3}{3}\right]_0^1 = \frac{1}{3}$
The value of the integral $\int_{-2}^{0} \frac{dx}{\sqrt{12 - x^2 - 4x}}$ is
1
2
3
4
Solution
$12 - x^2 - 4x = 16 - (x+2)^2$
Let $u = x+2$
Limits: $x=-2 \to u=0$, $x=0 \to u=2$
$\int_{0}^{2} \frac{du}{\sqrt{16 - u^2}}$
$= \sin^{-1}\left(\frac{u}{4}\right)\Big|_0^2$
$= \sin^{-1}\left(\frac{2}{4}\right) = \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}$
The value of
$ \displaystyle \int_0^{\infty} \frac{y^x,dx,dy}{x^2+y^2} $
is
1
2
3
4
Solution
Using standard double integral result over first quadrant:
$ \int_0^\infty \int_0^\infty \frac{dx,dy}{x^2+y^2} = \frac{\pi}{4} $
The value of
$ \iiint_V z,dx,dy,dz $
where $V$ is cylinder bounded by
$z=0,\ z=1,\ x^2+y^2=4$
1
2
3
4
Solution
Volume element in cylindrical coordinates:
$ \int_0^1 z,dz \int_{r=0}^2 r,dr \int_0^{2\pi} d\theta $
$= \left[\frac{1}{2}\right] \cdot \left[2\right] \cdot (2\pi)$
$= 2\pi$
If the Trapezoidal rule with interval $[0,1]$ is exact for approximating the integral
$\int_{0}^{1} (x^3 - cx^2)\,dx$, then the value of $c$ is
1
2
3
4
Solution
Actual integral:
$\int_{0}^{1} (x^3 - cx^2)\,dx = \left[\frac{x^4}{4} - \frac{cx^3}{3}\right]_0^1 = \frac{1}{4} - \frac{c}{3}$
Trapezoidal rule with one interval:
$T = \frac{1}{2}[f(0) + f(1)]$
$f(0) = 0$
$f(1) = 1 - c$
$T = \frac{1}{2}(1 - c)$
Since exact:
$\frac{1}{4} - \frac{c}{3} = \frac{1}{2}(1 - c)$
Solving:
$\frac{1}{4} - \frac{c}{3} = \frac{1}{2} - \frac{c}{2}$
Multiply by 12:
$3 - 4c = 6 - 6c$
$2c = 3$
$c = 1.5$
AMU MCA
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
AMU MCA
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
Ask Your Question or Put Your Review.
