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Let \(I=\int\limits_0^1\cfrac{tan^-1x}{1+x^2}dx\)

Let tan^-1x=t

\(\Rightarrow\cfrac{1}{1+x^2}dx=dt.\)

Also, when x=0, t=0

and when x=1,t=\(\cfrac{\pi}{4}\)

Hence,

\(I= \int\limits_0^{\pi/4}t\,dt=\cfrac{1}{2}t^2|_0^{\frac{\pi}{4}}=\cfrac{\pi^2}{32}\)

\(\int\limits_2^43dx=3[x]\)

=3[4-2]

=6

Correct option is: (a) 1 sq unit

Correct option is: (a) 12 sq units

Correct option is: (a) 128/3 sq units

Correct option is: (a) 16a²/3 sq units 

Correct option is: (a) 25π/3 sq units 

Correct option is: (a) 1/3 log 2 sq units

Correct option is: (d) 56/3 sq units

Correct option is: (b) πab/4 - ab/2  sq units 

Correct option is: (b) 9 sq units

Correct option is: (d) 1/3 sq unit

Let, sin x + cos x = t 

⇒ cos x – sin x dx = dt 

At x = 0, t = 1 

At x = π/2, t = 1

\(y=\int\limits_0^{1}-logt\,dt\)

We know that when upper and lower limit in definite integral is 

equal then value of integration is zero. 

So, y = 0

Let I =  \(\int\limits_{0}^{1}\cfrac{2\text x}{1+\text x^2}d\text x \)

Substituting 1+x² = t

⇒ 2x dx=dt

Also, When x = 0, t=1 and x=1, t = 2

We get

I = \(\int\limits_{0}^{1}\cfrac{1}tdt\)

= log_et \(|^2_1\) = log_e 2 - log_e1

= log_e2 (Since log_e1 = 0)

\(=\int\limits_1^2 log(x)dx=xlog(x)-(x)\) 

\(=2log(2)-(2)-1log(1)+(1)\)

= 2log(2)-1

Correct option is: (a) \(\cfrac{4-\pi}{2}\)

Correct option is: (a) \(\frac{1}{3} log (\frac{208}{189})\)

Correct option is: (c) √e(√e − 1) 

Correct option is: (d) 4√2

Correct option is: (c) 5π/256

Given MC = f'(x) = 10 + 6x – 6x², C = 125, TC = ? 

TC = C(x) = ∫MCdx = ∫(10 +6x – 6x²)dx = 10x + \(\frac{6x^2}{2} - \frac{6x^3}{3}\)+ C 

Total cost = 10x + 3x² – 2x³ + 125.

\(\int\limits_0^{\pi/6}sec^2x\,dx=[tanx]\)

\(=[tan(\frac{\pi}{6})-tan\,0]\)

\(=\cfrac{1}{\sqrt{3}}\)

\(\int_0^1x^2dx\) = \(\Big[\frac{x^3}{3}\Big]_0^1 = \frac{1}{3}\)

To find: \(\int\cfrac{e^x}{(e^{2x}+1)}dx\)

Formula Used: \(\int\cfrac{dx}{1+x^2}=tan^{-1}x\)

Let y = e^x … (1) 

Differentiating both sides, we get '

dy = e^x dx 

Substituting in given equation,

\(\Rightarrow\) \(\int\cfrac{dy}{y^2+1}\)

⇒ tan^-1 y

From (1), 

⇒ tan^-1 (e^x ) 

Therefore,

\(\int\cfrac{e^x}{(e^{2x}+1)}dx=tan^{-1}(e^x)+c\)

\(\int\limits_0^1\frac{dx}{1+x^2}\) = \([tan^{-1}x]^1_0\)

= tan^-11 - tan^-10

= \(\frac{\pi}{4}-0\) 

= \(\frac{\pi}{4}\)

Correct answer: D. 0

f(x)=tan x 

f(-x) =tan(-x) 

=-tan x 

hence the function is odd, 

therefore, I=0

\(\int\limits_{-4}^{-1}\cfrac{dx}{x}=-[logx]\)

=[log(-1)-log(-4)] 

=-[log(-4)-log(-1)]

=\(-\left[log\left(\cfrac{-4}{-1}\right)\right]\)

=-log 4

 =\(\int\limits_{\pi/6}^{\pi/3}\)tan²x + 2tanx cot x + cot²x dx

recall: sec²x - tan²x = 1, cosec²x - cot²x = 1

= \(\int\limits_{\pi/6}^{\pi/3}\)sec²x -1 + 2 + cosec²x - 1 dx

= \(\int\limits_{\pi/6}^{\pi/3}\) sec²x + cosec²x dx

Integral sec²x is tan x and integral of cosec²x = - cot x

= [tan x]\(\cfrac{\frac{\pi}3}{\frac{\pi}6}\) - [cot x]\(\cfrac{\frac{\pi}3}{\frac{\pi}6}\)

Tan 30 = cot 60 = \(\cfrac{1}{\sqrt3}\)

Tan 60 = cot 30 = √3

= Tan 60 - cot 60 - {Tan 30 - cot 30}

= \(\sqrt3-\cfrac{1}{\sqrt3}-(\cfrac{1}{\sqrt3}-\sqrt3)\)

= \(\cfrac{4}{\sqrt3}\)

Let I = \(\int\limits_{0}^{1}\cfrac{1}{\text x^2+1}d\text x\)

Substituting x=tanθ ⇒ dx=sec²θdθ (By differentiating both sides)

Also, when x=0, θ=0 and x=1 θ = \(\cfrac{\pi}4\)

Since sec²θ=1 + tan²θ

We get I = \(\int\limits_{0}^{\pi/4}d\theta\)

= \(\cfrac{\pi}4\)

Correct option is: (d) I₁ = I₂

Correct option is: (a) 0 

Let

I =  \(\int\limits_{-1}^{1} \) x | x | dx

|x|= -x, if x <0

And |x|=x, if x ≥ 0

Therefore f(x)=x|x|=-x², if x<0

And f(x)=x|x|=x², if x ≥ 0

Consider x≥0 ⇒ f(x)=x²

Then -x < 0 ⇒ f(-x) = -(-x)² = -f(-x)

Now Consider x<0 ⇒ f(x)=-x²

Then -x ≥ 0 ⇒ f(-x) =-(-x)²=x²= -f(x)

Hence f(x) is an odd function. An odd function is a function which satisfies the propertyf(-x) =-f(-x), ∀ x∈ Domain of f(x)

There is a property of integration of odd functions which states that

\(\int\limits_{-a}^af(x)dx=0 \) if f(x) is an odd function.

Therefore I =  \(\int\limits_{-1}^{1} \) x | x | dx = 0

Let I =  \(\int\limits_{-\pi/2}^{\pi/2} \) x cos²x dx

f(x)=xcos²x

f(-x)=(-x)cos²(-x)

=-xcos²x

=-f(x)

 Hence, f(x) is an odd function.

Since, \(\int\limits_{-a}^af(x)dx=0\) if f(x) is an odd function.

Therefore, I = 0.

Let I = \(\int\limits_{-\pi/2}^{\pi/2}\)sin³x dx

f(x)=sin³x

 f(-x)=sin³(-x)=-sin³x

Hence, f(x) is an odd function.

Since. \(\int\limits_{-a}^af(x)dx=0\) if f(x) is an odd function.

Therefore, I = 0