Definite Integrals MCQ Quiz - Objective Question with Answer for Definite Integrals - Download Free PDF

Analysis:

Consider \(I = \mathop \smallint \limits_0^{\pi /2} \sqrt {\sin \theta } {\cos ^5}\theta {\rm{ }}d\theta \)

Put sin θ = t

cos θ dθ = dt

If θ = 0 to \(\theta = \frac{\pi }{2}\) then t = 0 to t = 1

Now, \(I = \mathop \smallint \limits_{\theta = 0}^{\pi /2} \sqrt {\sin \theta } \cos \theta \cdot {\left( {1 - {{\sin }^2}\theta } \right)^2}d\theta \)

\(I = \mathop \smallint \limits_{t = 0}^1 \sqrt t {\left( {1 - {t^2}} \right)^2}dt\)

\(I = \mathop \smallint \limits_{t = 0}^1 \sqrt t \left( {1 + {t^4} - 2{t^2}} \right)dt\)

\(I = \left( {\frac{{{t^{\frac{3}{2}}}}}{{\frac{3}{2}}} + \frac{{{t^{\frac{{11}}{2}}}}}{{\frac{{11}}{2}}} - \frac{{2{t^{\frac{7}{2}}}}}{{\frac{7}{2}}}} \right)_0^1\)

\(I = \frac{2}{3} + \frac{2}{{11}} - \frac{4}{7}\)

Concept:

Trigonometric Ratio Fundamental Identities

\(\sec x ={1\over \cos x}\)

Integration by parts

when u and v are functions of x.

\(\smallint u× v~dx = u\smallint vdx - \smallint \left[ {\frac{{du}}{{dx}}\smallint vdx} \right]dx\)

Integral of standard function 

\(\smallint \sec^2(ax +b) dx=\tan (ax+b)\)

\(\smallint\tan x dx= log |\sec x|=-log~cos x\)

The derivative of standard function

\(\frac{{d}}{{dx}}(x^n)=nx^{n-1}\)

\(\frac{{d}}{{dx}}(x)=1\)

Calculation:

Given:

we have \(x\over cos^2 x\)

where u = x and \(v ={1\over cos^2 x}= sec^2 x\)

Integration by parts

when u and v are functions of x.

Integration by parts

when u and v are functions of x.

\(\smallint u× v~dx = u\smallint vdx - \smallint \left[ {\frac{{du}}{{dx}}\smallint vdx} \right]dx\)

\(\smallint x× sec^2 x~dx = x\smallint sec^2 x~dx - \smallint \left[ {\frac{{d}}{{dx}}x\smallint sec^2x~dx} \right]dx\)

\(\smallint x× sec^2 x~dx = x \tan x~dx - \smallint \tan x dx\)

∫x × sec² x dx = x tan x - ∫ tan x dx

∫x × sec2 x dx = x tan x - (-log (cos x))

∫ x × sec2 x dx = x tan x + log (cos x)

Concept:

Arc length or curve length is the distance between two points along the section of the curve. Determining the length of an irregular section of the arc is termed as rectification of the curve.

The length of the curve y = f(x) from x = a to x = b is given as:

\(l= \mathop \smallint \limits_{{\rm{x}} = {\rm{a}}}^{{\rm{x}} = {\rm{b}}} \sqrt {1 + {{\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)}^2}} {\rm{dx}}\)

or,

If the curve is parametrized in the form x = f(t) and y = g(t) with the parameter t going from a to b then

\(l = \mathop \smallint \limits_{{\rm{t}} = {\rm{a}}}^{{\rm{t}} = {\rm{b}}} \sqrt {{{\left( {\frac{{{\rm{dx}}}}{{{\rm{dt}}}}} \right)}^2} + {{\left( {\frac{{{\rm{dy}}}}{{{\rm{dt}}}}} \right)}^2}} {\rm{dt\;}}\)

Calculation:

\({\rm{y}} = \frac{2}{3}{\rm{\;}}{{\rm{x}}^{3/2}}\)

\(\therefore \frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \frac{2}{3} \times \frac{3}{2}{\rm{\;}}{{\rm{x}}^{\frac{1}{2}}}\)

\({\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)^2} = {\rm{x}}\)

Now, the arc length(l) is

\(l= \mathop \smallint \limits_{{\rm{x}} = 0}^{{\rm{x}} = 1} \sqrt {1 + {\rm{x}}} {\rm{\;dx}}\)

\( = \left| {\frac{{{{\left( {1 + {\bf{x}}} \right)}^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right|_0^1 \)

\(= \left( {\frac{2}{3} \times {2^{\frac{3}{2}}}} \right) - \left( {\frac{2}{3}} \right) = 1.21\)

Explanation:

The given integral is an improper integral of 1^st kind.

\(I = \mathop \smallint \limits_2^∞ \frac{{\left( {1/x} \right)}}{{\log x}}dx\)

\(I = \left[ {\log \left( {\log x} \right)} \right]_2^∞ \)

I = log [log (∞)] – log [log (2)]

∴ I = ∞

Given integral is divergent and diverges to ∞

Additional Information

An improper integral of first kind is when integral limits have -∞ or +∞ or both.

An improper integral of second kind is when integral limits are finite but function is infinite at some value between those limits.

Concept:

Using Integration by parts using ILATE

\(\smallint f\left( x \right)g\left( x \right)dx = f\left( x \right)\smallint g\left( x \right)\;dx - \smallint [\left( {f'\left( x \right).\smallint g\left( x \right)\;dx} \right]dx\)

Calculation:

\({\rm{I}} = \mathop \smallint \nolimits_0^{{\rm{\pi }}/2} {{\rm{x}}^2}\sin {\rm{x\;dx}}\)

\({\rm{f}}\left( {\rm{x}} \right) = {{\rm{x}}^2}{\rm{\;and\;g}}\left( {\rm{x}} \right) = {\rm{sinx}}\)

\({\rm{I}} = \left[ { - {{\rm{x}}^2}{\rm{cosx}} - \smallint \left( { - 2{\rm{x\;cosx}}} \right)} \right]_0^{{\pi }/{2}}\;\)

\({\rm{I}} = \left[ { - {{\rm{x}}^2}{\rm{cosx}} + 2 \times \left( {{\rm{xsinx}} - \smallint {\rm{sinx\;}}} \right)} \right]_0^{{\pi }/{2}}\)

\({\rm{I}} = \left[ { - {{\rm{x}}^2}{\rm{cosx}} + 2 \times \left( {{\rm{xsinx}} + {\rm{cosx\;}}} \right)} \right]_0^{{\pi }/{2}}\)

\({\rm{I}} = \left[ {0 + 2\;\left( {\frac{\pi }{2}\;\left( 1 \right) + 0} \right)} \right] - \left[ {0 + 2\;\left( {0\; + 1} \right)} \right]\)

\({\rm{I}} = {\rm{\pi }} - 2\)

Concept:

We know that,

By Parts method

\(\smallint u \cdot v \cdot dx = u \cdot \smallint v \cdot dx - \smallint \left[ {\frac{{du}}{{dx}}\smallint v \cdot dx} \right]dx\)

Where, u, v should follow the ILATE sequence.[I= Inverse, L= Logarithmic, A= Algebraic, T= Trigonometric, E= Exponential terms]

Calculation:

Given:

From the given Equation, \(\mathop \smallint \limits_1^e √ x \ln \left( x \right)dx\)

u = ln(x), v = √x

Now,

\(\smallint u \cdot v \cdot dx = u \cdot \smallint v \cdot dx - \smallint \left[ {\frac{{du}}{{dx}}\smallint v \cdot dx} \right]dx\)

\(\mathop \smallint \limits_1^{\rm{e}} {\rm{lnx}} \cdot √ {\rm{x}} {\rm{dx}} = {\rm{lnx}} \cdot \mathop \smallint \limits_1^{\rm{e}} √ {\rm{x}} {\rm{dx}} - \mathop \smallint \limits_1^{\rm{e}} \left[ {\frac{{{\rm{du}}}}{{{\rm{dx}}}} \cdot {\rm{\;}}\smallint √ {\rm{x}} {\rm{dx}}} \right]{\rm{dx}}\)  

\(\mathop \smallint \limits_1^{\rm{e}} {\rm{lnx}} \cdot √ {\rm{x}} {\rm{dx}}= \left[ {\ln \left( x \right) \times \frac{{{x^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right]_1^e - \smallint \left[ {\frac{1}{x} \times \frac{{{x^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right]dx\)

 \(\mathop \smallint \limits_1^{\rm{e}} {\rm{lnx}} \cdot √ {\rm{x}} {\rm{dx}}= \left[ {\ln \left( x \right) \times {x^{\frac{3}{2}}} \times \frac{2}{3} - \frac{4}{9} \times {x^{\frac{3}{2}}}} \right]_1^e\)

∴  \(\mathop \smallint \limits_1^e √ x \ln \left( x \right)dx\)  \(= \frac{2}{9}√ {{e^3}} + \frac{4}{9}\)     

Explanation

Given,

 Function f(x) = x e^|x|

 Integral is -1 to 1.

If f(-x) = f(x) then the function is said to be even function

 If f(-x) = - f(x) then the function is said to be odd function.

 f(-x) = -x e^|-x| = -x e|x| = - f(x)

 ∴ The given function is an odd function.

 For an odd function:

 \(\mathop \smallint \nolimits_{ - a}^a x\;{f(x)}dx\) = 0

For a even function

 \(\mathop \smallint \nolimits_{ - a}^a x\;{f(x)}dx\) = 2 × \(\mathop \smallint \nolimits_{ 0}^a x\;{f(x)}dx\)

 Now, as the function is odd

 \(\mathop \smallint \nolimits_{ - 1}^1 x\;{e^{\left| x \right|}}dx\) = 0

Explanation:

\(I=\mathop \smallint \limits_{ - 1}^1 {e^{\left| x \right|}}dx.\)

\(I= \mathop \smallint \limits_{ - 1}^0 {e^{ - x}}dx + \mathop \smallint \limits_0^1 {e^x}dx\)

\(I= \left[ { - {e^{ - x}}} \right]_{-1}^0 + \left[ {{e^x}} \right]_0^1\)

I = [-e^-0 + e¹] + [e¹ - e⁰]

I = -1 + e¹ + e - 1

Concept:

\(x = \cos \left( {\frac{{\pi u}}{2}} \right),\;y = \sin \left( {\frac{{\pi u}}{2}} \right)\)

\({x^2} + {y^2} = {\cos ^2}\left( {\frac{{\pi u}}{2}} \right) + {\sin ^2}\left( {\frac{{\pi u}}{2}} \right) = 1\)

So it represents an equation of circle in x-y plane.

Given 0 ≤ u ≤ 1

So, 0 ≤ x ≤ 1,  0 ≤ y ≤ 1

i.e., 0 ≤ θ ≤ π/2

So we get a quarter circle in x-y plane and by revolving it 360°, we get a hemisphere.

Area of hemi-sphere = 2π(1)² = 2π

Concept:

Volume of the solid of rotation obtained by rotating the shaded area by 360° around the x-axis is asked,

there is a direct relation for this;

 \({\forall _1} = \smallint \pi {y^2}dx\)…1)

Calculation:

Given area;

Using (1); Volume of solid of rotation can be calculated by:

\({\forall _1} = \mathop \smallint \nolimits_0^1 \pi xdx\)

\( = \pi \mathop \smallint \nolimits_0^1 xdx = \pi \left\{ {\frac{{{x^2}}}{2}} \right\}_0^1 \)

\(= \frac{\pi }{2}\left\{ {1 - 0} \right\}\)

\({\forall _1} = \frac{\pi }{2}units;\)

Key points:

In the given problem, the volume is generated by revolving the area by 360° about the x-axis.

But if rotation/revolution is about the y-axis, then the volume of solid of rotation is calculated by:

\({\forall _2} = \smallint \pi {x^2}dy\) …2)

So, always be careful about which axis rotation is asked.

Depending upon that, you should use either 1) or 2).

f(x) = x – [x]

for 0.25 < x < 1, f(x) = x

for 1 < x < 1.25, f(x) = x – 1

\(\mathop \smallint \limits_{0.25}^{1.25} f\left( x \right)dx = \mathop \smallint \limits_{0.25}^1 x\;dx + \mathop \smallint \limits_1^{1.25} \left( {x - 1} \right)dx\)

\(= {\left[ {\frac{{{x^2}}}{2}} \right]_{0.25}}^1 + {\left[ {\frac{{{x^2}}}{2} - x} \right]_1}^{1.25}\)

\(= \frac{1}{2}\left[ {1 - \frac{1}{{16}}} \right] + \left[ {\left( {\frac{{25}}{{32}} - \frac{5}{4}} \right) - \left( {\frac{1}{2} - 1} \right)} \right]\)

\(= \frac{1}{2}\left[ {\frac{{15}}{{16}}} \right] + \left[ {\frac{{25}}{{32}} - \frac{5}{4} + \frac{1}{2}} \right]\)

\(= \frac{{15 + 25 - 40 + 16}}{{32}} = 0.5\)

Given x as non- zero,

\(af\left( x \right) + bf\left( {\frac{1}{x}} \right) = \frac{1}{x} - 25\)       ---(1)

Consider x as 1/x

\(af\left( {\frac{1}{x}} \right) + bf\left( {\frac{1}{{\frac{1}{x}}}} \right) = x - 25\)

\(af\left( {\frac{1}{x}} \right) + bf\left( x \right) = x - 25\)       ---(2)

Multiply equation 1 by a and 2 by b and subtract both

\({a^2}f\left( x \right) + abf\left( {\frac{1}{x}} \right) - abf\left( {\frac{1}{x}} \right) - {b^2}f\left( x \right) = \frac{a}{x} - 25a - bx + 25b\)

\({a^2}f\left( x \right) - {b^2}f\left( x \right) = \frac{a}{x} - 25a - bx + 25b\)

\(f\left( x \right) = \frac{a}{{x\left( {{a^2} - {b^2}} \right)}} - \frac{{bx}}{{{a^2} - {b^2}}} - \frac{{25\left( {a - b} \right)}}{{{a^2} - {b^2}}}\)

\(\mathop \smallint \limits_1^2 f\left( x \right)dx = \frac{1}{{{a^2} - {b^2}}}\left[ {a\left( {\ln 2 - 25} \right) + \frac{{47b}}{2}} \right]\)

Concept:

\(\int_a^b f(x)dx = f(a + b - x)dx\)

Calculation:

⇒ Let, I = \(\int \limits_0^\frac{π}{2} \frac{f(x)}{f(x) + f{(\frac{π}{2} - x)}}dx\)    ----- equation(1)

⇒ I = \(\int \limits_0^\frac{π}{2} \frac{f(\frac{π}{2}- x )}{f(\frac{π}{2} - x) + f{(\frac{π}{2} -(\frac{π}{2} - x))}}dx\)

⇒ I = \(\int \limits_0^\frac{π}{2} \frac{f(\frac{π}{2}- x )}{f(x) + f{(\frac{π}{2} - x)}}dx\)     ---- equation(2)

On adding equation (1) and (2)

⇒ 2I = \(\int \limits_0^\frac{π}{2} \frac{ f(x) + f(\frac{π}{2}- x )}{f(x) + f{(\frac{π}{2} - x)}}dx\)

⇒ 2I = \(\int \limits_0^\frac{π}{2} dx\)

⇒ 2I = \(\frac{π}{2}\)

⇒ I = \(\frac{π}{4}\)

∴ The value of \(\int_{\rm{0}}^{\frac{{\rm{\pi }}}{{\rm{2}}}} {\frac{{{\rm{f(x)}}}}{{{\rm{f(x) \,+\, f}}\left( {\frac{{\rm{\pi }}}{{\rm{2}}}\,{\rm{ - }}\,{\rm{x}}} \right)}}{\rm{dx}}} \) is \(\frac{π}{4}\)

Concept:

Following steps to solve the equation \(\mathop \smallint \nolimits_0^{2\pi } \left| {x\;sin\;x} \right|dx = k\pi ,\)

1. To remove the modulus
2. To keep sin x positive in the interval 0 to π to 2π and to keep the sin x negative in the interval.This is because x in the above equation is always positive but the value sinx changes in the two mentioned intervals.

Explanation:

Solving the equation as per the steps,

\(\mathop \smallint \nolimits_0^{2\pi } \left| {x\;sin\;x} \right|dx = \mathop \smallint \nolimits_0^\pi xsinxdx + \left( { - \mathop \smallint \nolimits_\pi ^{2\pi } xsinxdx} \right)\)

\(= \mathop \smallint \nolimits_0^\pi xsinxdx - \mathop \smallint \nolimits_\pi ^{2\pi } xsinxdx\)

Keeping u = x, du = dx, dv = sinxdx, so v = - cosx,

\(= \mathop \smallint \nolimits_0^\pi udv = uv - \mathop \smallint \nolimits_0^\pi vdu\)

\( = \mathop \smallint \nolimits_0^\pi xsinxdx = \left[ { - xcosx} \right]_0^\pi + \mathop \smallint \nolimits_0^\pi cosxdx\;\)

\( = \pi + \left[ {sinx} \right]_0^\pi \)

Now, repeating the same with \(- \mathop \smallint \nolimits_\pi ^{2\pi } xsinxdx\), we get -3

Hence, π - (-3π) = 4π

Therefore, k = 4

Concept:

Use integration by parts for solving this problem.

Calculation:

\(\mathop{\int }_{1}^{e}\left( logx~x \right)dx\)

\(=\left\{ \log x.\frac{{{x}^{2}}}{2} \right\}_{1}^{e}-\mathop{\int }_{1}^{e}\frac{1}{x}.\frac{{{x}^{2}}}{2}dx\) 

\(=\left\{ \frac{{{x}^{2}}\log x}{2} \right\}_{1}^{e}-\frac{1}{2}\mathop{\int }_{1}^{e}xdx\) 

\(=\left\{ \frac{{{x}^{2}}\log x}{2} \right\}_{1}^{e}-\left\{ \frac{{{x}^{2}}}{4} \right\}_{1}^{e}\) 

\(=\left\{ \frac{{{e}^{2}}\log e}{2}-\frac{\log 1}{2} \right\}-\left\{ \frac{{{e}^{2}}}{4}-\frac{1}{4} \right\}\) 

\(=\frac{{{e}^{2}}}{2}-\frac{{{e}^{2}}}{4}+\frac{1}{4}\) 

\(\Rightarrow \frac{{{e}^{2}}+1}{4}=2.097\)