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This mock test of Test: Problems On Integration - 1 for JEE helps you for every JEE entrance exam.
This contains 10 Multiple Choice Questions for JEE Test: Problems On Integration - 1 (mcq) to study with solutions a complete question bank.
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QUESTION: 1

Evaluate:

Solution:

I = ∫√(x² + 5x)dx

= ∫√(x² + 5x + 25/4 - 25/4)

= ∫√{(x + 5/2)² - (5/2)²}

={1/2(x+5/2)(√x² + 5x)} - {25/8 log{(x + 5/2)+√x²+ 5x}}

= {(2x + 5)/4 (√x² + 5x)} - {25/8 log{(x + 5/2)+√x²+ 5x}}

Thus, option D is correct...

QUESTION: 2

Evaluate:

Solution:

sin^{2}x = 1 - cos^{2}x

∫sinx(sin^{2}x - 3cos^{2}x + 15)dx

Put cos^{2}x = t

∫sinx(1 - cos^{2}x - 3cos^{2}x + 15)dx

= ∫sinx (16 - 4cos^{2}x)dx

Put t = cosx, differentiate with respect to x, we get

dt/dx = -sinx

= - ∫ [(16 - 4t^{2})]^{1/2}dt

= -2 ∫ [(2)^{2} - (t)^{2}]^{½}

= -2{[(2)^{2} - (t)^{2}]^{½ }+ 2sin^{-1}(t/2)} + c

= - cosx {[4 - (cos)^{2}x]^{½} - 4sin^{-1}(cosx/2)} + c

QUESTION: 3

Evaluate:

Solution:

Let's apply the integral substitution,

substitute

Now use the standard integral :

substitute back u=(x-2) and add a constant C to the solution,

QUESTION: 4

Evaluate:

Solution:

(x)^{½} (a - x)^{½} dx

= ∫(ax - x^{2})^{½} dx

= ∫{-(x^{2} - ax)^{½}} dx

= ∫{-(x^{2} - ax + a^{2/4} - a^{2/4})^{½}} dx

= ∫{-(x - a/2)^{2} - a^{2/4}} dx

= ∫{(a/2)^{2} - (x - a/2)^{2}} dx

= ½(x - a/2) {(a/2)^{2} - (x - a/2)^{2}} + (a^{2/4}) (½ sin^{-1}(x - a)/a^{2})

= {(2x - a)/4 (ax - x^{2})^{½}} + {a^{2/8} sin^{-1}(2x - a)/a} + c

QUESTION: 5

is equal to

Solution:

QUESTION: 6

Evaluate:

Solution:

QUESTION: 7

Solution:

QUESTION: 8

Evaluate:

Solution:

Apply Trig Substitution: x=4sin(u)

QUESTION: 9

Evaluate:

Solution:

Let y = e^{x}

dy/dx = e^{x} dx

= ∫(9 - y^{2})^½ dy

= ∫[(3)^{2} - (y)^{2}]^{½} dy

Apply [(a)^{2} - (x)^{2}] formula

⇒ y/2[(3)^{2} - (y)^{2}]^{½} + [(3)^{2}]/2 sin^{-1}(y/3) + c

= e^{x}/2[(3)^{2} - (y)^{2}]^{½} + [9]/2 sin^{-1}(e^{x}/3) + c

QUESTION: 10

The integral of the function w.r.t. x is:

Solution:

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