Formula used:

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

Calculation:

Let,

\(I=\frac{1}{c}\displaystyle \int_{ac}^{bc}f\left(\frac{x}{c}\right)dx\)   ---(1)

Let x/c = t

⇒ dx = c dt

When x = ac ⇒ t = a

When x = bc ⇒ t = b

Hence, from equation (1)

\(I=\frac{1}{c}\displaystyle \int_{ac}^{bc}f\left(\frac{x}{c}\right)dx\)

\(\Rightarrow I=\frac{1}{c}\times c\displaystyle \int_{a}^{b}f\left(t\right)dt\)

\(\Rightarrow I=\displaystyle \int_{a}^{b}f\left(t\right)dt\)

\(\therefore I =\displaystyle \int_{a}^{b} f(x)dx\)

Concept:

• sin² x + cos² x = 1
• sin 2A = 2 sin A cos A
• ∫ cos x = sin x

• ∫ sin x = -cos x

Calculation:

\(∫_{0}^{\pi /3}\sqrt{1+\sin 2x}\, dx\)

\(\Rightarrow ∫_{0}^{\pi /3}\sqrt{(\sin ^2x\, +\, \cos ^2x\, +\, 2\sin x\, \cos x})\, dx\)

\(\Rightarrow ∫_{0}^{\pi /3}(\cos x\, +\,\sin x)\, dx\)

\(\Rightarrow \left [ \sin x \, -\, \cos x \right ]_{0}^{\pi /3}\)

\(\Rightarrow \left [ \sin \frac{\pi }{3}\, -\, \cos \frac{\pi }{3} \right ]-(\sin 0\, -\, \cos 0) \)

\(\Rightarrow \left [ \frac{\sqrt{3} }{2}\, -\, \frac{1 }{2} \right ]-(0\, -\, 1)\)

\(\Rightarrow \left [ \frac{\sqrt{3}-1 }{2} \right ]+ 1\)

\(\Rightarrow \frac{\sqrt{3}+1}{2}\)

Hence, \(∫_{0}^{\pi /3}\sqrt{1+\sin 2x}\, dx=\frac{\sqrt{3}+1}{2}\)

Concept Used:

• \(\rm \int cos x \space dx = sin x \)
•  \(e^{ln(f(x))} = f(x)\)

Calculation:

Let I = \(\rm \int^\frac{\pi}{2}_0 e^{ln(cos x)} dx\)

According to the concept used

⇒ I = \(\rm \int^\frac{\pi}{2}_0 (cos x) dx\)

⇒  \(\rm [sinx]_{0}^{\frac{\pi }{2}}\)

⇒  \(\rm sin \frac{\pi }{2} - sin 0\)

⇒ 1 - 0 = 1

∴ The value of the integral \(\rm \int^\frac{\pi}{2}_0 e^{ln(cos x)} dx\)  is 1.

Formula used :

Property of definite integral = \(\displaystyle \int_{-k}^{k}f(x)\:dx = 2\int _{0}^{k} f(x)dx\) .... if f(-x) = f(x) or it is an even function      ---- (1)

Calculations :

Now, \(\int_{\frac{-π}{2}}^{\frac{π}{2}}|sin\space x|dx\)

= \(\int_{\frac{-π}{2}}^{0}(-sin\space x)dx\) + \(\int_{0}^{\frac{π}{2}}sin\space xdx\)

= \([\cos x]_{-\frac{π}{2}}^0 + [-\cos x]_0^{\frac{π}{2}}\)

= [ 1 - cos (-π/2)] - [cos(π/2) - cos 0]

= [1 - cos(π/2)] - [0 - 1]  (∵ cos (- x) = cos x)

= [1 - 0 ] - [- 1]

= 1 + 1

= 2

∴ The value of \(\displaystyle \int_{-π/2}^{π/2}|\sin x|\:dx \space is \ 2\).

Concept:

• \(\rm \int \frac{dx}{x^{2}+a^{2}}=\frac{1}{a}tan^{-1}\frac{x}{a}+C\)
• \(\rm \int_{0}^{x}{f(x) dx}=F(x)-F(0)\), where F(x) is the anti-derivative of f(x).

Calculation:

Given: \(\rm \int_{0}^{1} \frac{dt}{t^{2}+25}=\frac{1}{5}(tan^{-1}{b}-tan^{-1}{a)}\)

Using the formula, \(\rm \int \frac{dx}{x^{2}+a^{2}}=\frac{1}{a}tan^{-1}\frac{x}{a}+C\)

\(\rm \Rightarrow \int_{0}^{1} \frac{dt}{x^{2}+25}=\int_{0}^{1} \frac{dt}{x^{2}+5^{2}}= \frac{1}{5}(tan^{-1}\frac{1}{5}-tan^{-1}\frac{0}{5})=\frac{1}{5}(tan^{-1}\frac{1}{5}-tan^{-1}0)\)---(1)

∵ It is given that, \(\rm \int_{0}^{1} \frac{dt}{t^{2}+25}=\frac{1}{5}(tan^{-1}{b}-tan^{-1}{a)}\)------(2)

On comparing (1) and (2) we get, a = 0, b = 1/5.

\(\rm \Rightarrow a+b=\frac{1}{5}+0=\frac{1}{5}\)

Hence, the correct answer is option 3.

Formula used:

Property 1:

 \(\int_{0}^{2p}f(a)da= \int_{0}^{p}f(a)da+\int_{0}^{p}f(2p-a)da\)

Property 2:

 \(\int_{a}^{b}f(x)dx= \int_{a}^{b}f(a +b- x)dx\)

Calculation:

Given that

f(2a - x) = f(x)          -----(1)

\(\displaystyle\int^{a}_0f(x)dx= \lambda\)     ------(2)

Using the property (1)

\(\int_{0}^{2a}f(x)dx = \int_{0}^{a}f(x)dx + \int_{0}^{a}f(2a - x)dx\)   

From equation (1)  

\(⇒ \int_{0}^{2a}f(x)dx =\int_{0}^{a}f(x)dx + \int_{0}^{a}f(x)dx\)   

From equation (2)   

⇒  \(\displaystyle\int^{2a}_0f(x)dx\) =  λ + λ                                        

⇒  \(\displaystyle\int^{2a}_0f(x)dx\) = 2λ 

Hence option 1 is correct.

CONCEPT:

Fundamental Theorem of Calculus:

Newton-Leibniz formula

\(\rm \frac{d}{dx} \left[ \int_{g(x)}^{h(x)} f(t) dt \right] = f(h(x)) . h'(x) - f(g(x)) g'(x)\)

CONCEPT:

Given:

\(I = \rm \frac{d}{dx} \int_{\sin x}^{\cos x} \frac{1}{t} dt\)

Using, the Newton-Leibniz formula,

\(I = = \frac{1}{ \cos x} ( - \sin x)-\frac{1}{sin\ x}(\cos x)\)

⇒ I = - tan x - cot x

So, the correct answer is option 3

Concept:

• \(\smallint {\rm{f''}}\left( {\rm{x}} \right){\rm{dx\;}} = {\rm{f'}}\left( {\rm{x}} \right) + {{\rm{C}}_1}\)
• \(\smallint {\rm{f'}}\left( {\rm{x}} \right){\rm{dx\;}} = {\rm{f}}\left( {\rm{x}} \right) + {{\rm{C}}_2}\)

Calculation:

Given f’’(x) = g’’(x)

Integrating both sides, we get

\(\smallint {\rm{f''}}\left( {\rm{x}} \right){\rm{dx}} = \smallint {\rm{g''}}\left( {\rm{x}} \right){\rm{dx}}\)

⇒ f’(x) = g’(x) + C

Using f’(1) = 4 , g’(1) = 6

f'(1) = g'(1) + C

⇒ 4 = 6 + C

⇒ C = -2

⇒ f’(x) = g’(x) - 2

Integrating both sides, we get

\( \Rightarrow \smallint {\rm{f'}}\left( {\rm{x}} \right){\rm{dx}} = \smallint ({{\rm{g}}^{\rm{'}}}\left( {\rm{x}} \right) - 2){\rm{dx}}\)

⇒ f(x) = g(x) - 2x + A

Using f(2) = 3 and g(2) = 9

⇒ f(2) = g(2) - (2 × 2) + A

⇒ 3 = 9 - (2 × 2) + A

⇒ A = 3 – 9 + 4 = -2

⇒ f(x) = g(x) - 2x - 2

⇒ f(x) - g(x) = -2x - 2

So, f(x) – g(x) at x = 4:

f(4) - g(4) = -(2 × 4) – 2 = -10

Concept: 

The rules for integrating even and odd functions ,

If the function is even or odd and the interval is [-a, a], we can apply these rules:

• When f(x) is even ⇔ \(\rm\int_{-a}^{a}f(x)dx= 2 \int_{0}^{a}f(x)dx\)
• When f(x) is odd ⇔ \(\rm\int_{-a}^{a}f(x)dx= 0\) 

​Calculation: 

Let f (x) = x⁵ sin⁴ x 

f (-x) = (-x)⁵ sin⁴ (-x) = -x⁵ (-sin x )⁴ = -x⁵ sin⁴ x .

⇒ f(-x) = - f(x) 

So, f(x) is an odd function . 

We knoiw that, when f(x) is odd ⇔ \(\rm\int_{-a}^{a}f(x)dx= 0\) 

Hence , \(\rm \int_{-\pi/2}^{\pi/2} x^{5}\sin^{4}x dx\) = 0 .  

The correct option is 4 .

Given:

f(x) = 2x

Formula used:

\(\rm \frac{d}{dx} a^{x} = (In\: a).a^{x}\)

\(\rm \int 1.dx = x + C\)

\(\rm \int_{a}^{b} f(x) dx = F(b) - F(a)\), Where \(\rm \int f(x).dx = F(x) + C\)

Calculation:

We have f(x) = 2x     ----(i)

⇒ f^'(x) = 2x In(2)    ----(ii)

Now, We have to find the value of \(\rm \int^{10}_2\frac{f'(x)}{f(x)}dx\)

Form equation (i) and (ii), we get 

⇒ \(\rm \int^{10}_2 \frac{2^{x} In(2)}{2^{x}} dx\)

⇒ \(\rm \int^{10}_2 { In(2)} \space dx\)

⇒ \(\)In(2) [10 - 2]

⇒ 8 In(2)

∴​ The value of \(\rm \int^{10}_2\frac{f'(x)}{f(x)}dx\) is equal to 8 In(2).