Here we provide RD Sharma Class 12 Ex 20.3 Solutions Chapter 20 Definite Integrals for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 12 Ex 20.3 Solutions Chapter 20 Definite Integrals book pdf download. Now you will get step-by-step solutions to each question.

Textbook | NCERT |

Class | Class 12th |

Subject | Maths |

Chapter | 20 |

Exercise | 20.3 |

Category | RD Sharma Solutions |

**RD Sharma Class 12 Ex 20.3 Solutions Chapter 20 Definite Integrals**

**Question 1: Ten cards numbered 1 through 10 are placed in a box, mixed **

### Evaluate the following integrals:

### Question 1(i).

**Solution:**

We have,

I =

I =

Using additive property, we get

I =

I =

I = 8 + 6 – 2 – 3 + 24 + 20 – 6 – 10

I = 37

### Question 1(ii).

**Solution:**

We have,

I =

I =

Using additive property, we get

I =

I =

I =

I = 0 + 1 + 3 – π/2 + e

^{6}– e^{0}I = 3 – π/2 + e

^{6}

### Question 1(iii).

**Solution:**

We have,

I =

I =

Using additive property, we get

I =

I =

I =

I = 63/2 + 9 – 7/2 – 3 + 64 – 36

I = 56/2 + 34

I = 62

### Question 2.

**Solution:**

We have,

I =

We know that,

So, we get

I =

I =

I =

I = –2 + 4 – 8 – 8 + 8 + 8 – 2 + 4

I = 20

### Question 3.

**Solution:**

We have,

I =

We know that,

So, we get

I =

I =

I = 0 + 2 + 8 – 0

I = 10

### Question 4.

**Solution:**

We have,

I =

We know that,

So, we get

I =

I =

I = –1/4 + 1/2 + 1 – 1 + 1 + 1 – 1/4 + 1/2

I = 5/2

### Question 5.

**Solution:**

We have,

I =

We know that,

So, we get

I =

I =

I = –9/4 + 9/2 + 4 – 6 + 4 + 6 – 9/4 + 9/2

I = 25/2

### Question 6.

**Solution:**

We have,

I =

We know that,

So we get,

I =

I =

I =

I = 1/3 – 3/2 + 2 – [8/3 – 6 + 4 – 1/3 + 3/2 – 2]

I = 1/3 – 3/2 + 2 – 8/3 + 6 – 2 + 1/3 – 3/2

I = 1

### Question 7.

**Solution:**

We have,

I =

We know that,

So we get,

I =

I =

I = –1/6 + 1/3 – 0 + 27/2 + 3 – 1/6 – 1/3

I = 65/6

### Question 8.

**Solution:**

We have,

I =

We know that,

So, we get

I =

I =

I =

I = –2 + 4 + 18 – 12 + 18 + 12 – 2 + 4

I = 40

### Question 9.

**Solution:**

We have,

I =

We know that,

So we get,

I =

I =

I =

I = -1/2 + 1 + 2 – 2 + 2 + 2 – 1/2 + 1

I = 5

### Question 10.

**Solution:**

We have,

I =

We know that,

So we get,

I =

I =

I =

I = – 2 – 6 + 1/2 + 3

I = – 5 + 1/2

I = (-10 + 1)/2

I = -9/2

### Question 11.

**Solution:**

We have,

I =

We know that,

So we get,

I =

I =

I =

I = 1/2 – 0 – 0 + 1/2

I = 1

### Question 12.

**Solution:**

We have,

I =

We know that,

So we get,

I =

I =

I =

I = 1 + 1 + 1 – (–1)

I = 1 + 1 + 1 + 1

I = 4

### Question 13.

**Solution:**

We have,

I =

We know that,

So we get,

I =

I =

I =

I = 1 – 1/√2 – 1/√2 + 1

I = 2 – 2/√2

I = 2 – √2

### Question 14.

**Solution:**

We have,

I =

We know that,

So we get,

I =

I =

I =

I = – 25/2 + 25 + 2 – 10 + 32 – 40 – 25/2 + 25

I = 9

### Evaluate the following integrals:

### Question 15.

**Solution:**

We have,

I =

Since f(- x) = sin|- x| + cos|- x|

= sin |x| + cos |x|

= f(x)

So, f(x) is an even function.

Therefore, we get

I =

I =

I =

I = 2 (0 + 1 + 1 – 0)

I = 4

### Question 16.

**Solution:**

We have,

I =

We know,

So we get,

I =

I =

I =

I = -1/2 + 1 – 0 + 8 – 4 – 1/2 + 1

I = 5

### Question 17.

**Solution:**

We have,

I =

We know,

So we get,

I =

I =

I = 8 – 4 – 1/2 + 1 – (2 – 4 – 1/2 + 2) + 8 – 8 – 2 + 4 – (8 – 16 – 1/2 + 4)

I = 23/2

### Question 18.

**Solution:**

We have,

I =

We know,

So we get,

I =

I =

I = 25/2 – (2 – 4 – 25/2 + 10) – 2 + 4 + (-25/2 + 25)

I = 63/2

### Question 19.

**Solution:**

We have,

I =

We know,

So we get,

I =

I =

I = 8 – (2 – 4) + 8 – 8 – 2 + 4 – (8 – 16)

I = 20

### Question 20.

**Solution:**

We have,

I =

We know,

When –1 < x < 0,

|x + 1| + |x| + |x – 1| = x + 1 + (- x) + [-(x – 1)]

= 2 – x

And when 0 < x < 1,

|x + 1| + |x| + |x – 1| = x + 1 + x + [-(x – 1)]

= x + 2

And when 1 ≤ x ≤ 2,

|x + 1| + |x| + |x – 1| = x + 1 + x + x – 1

= 3x

So we get,

I =

I =

I =

I = – 1/2(4 – 9) + 1/2( 9 – 4) + 3/2(4 – 1)

I = 5/2 + 5/2 + 9/2

I = 19/2

### Question 21.

**Solution:**

We have,

I =

Now here,

f(- x) = (- x)e

^{|- x|}= – x e

^{|x|}= – f(x)

So, f(x) is an odd function.

Therefore we get,

I =

I = 0

### Question 22.

**Solution:**

We have,

I =

I =

As we know,

I =

I =

I =

I =

I =

I = -1/2(0 + π/4) + 1/4(0 + sin π/2) + 1/2 ( π/2 – 0) – 1/4(sin π – 0)

I = – π/8 + 1/4 (0 + 1) + π/8 – 1/4 (0 – 0)

I = π/8 + 1/4

### Question 23.

**Solution:**

We have,

I =

Now here,

f(π – x) = cos(π – x)|cos(π – x)|

= -cos x|-cos x|

= – cos x|cos x|

= – f(x)

So, f(x) is an odd function.

Therefore we get,

I =

I = 0

### Question 24.

**Solution:**

We have,

I =

Now here,

f(- x) = 2sin|- x| + cos|- x|

= 2sin|x| + cos|x|

= f(x)

So, f(x) is an odd function.

Therefore we get,

I =

I =

As we know,

I =

I =

I =

I = – 4(cos π/2 – cos 0) + 2(sin π/2 – sin 0)

I = –4 ( 0 – 1) + 2 (1 – 0)

I = 4 + 2

I = 6

### Question 25.

**Solution:**

We have,

I =

I =

I =

As π/2 ≤ x ≤ π, we get

=> –π ≤ –x ≤ –π/2

=> 0 ≤ π – x ≤ π/2

So, we get

I =

I = 1/2 (π

^{2}/4 – π^{2}/4) – 1/2( 0 – π^{2}/4)I = 0 + π

^{2}/8I = π

^{2}/8

### Question 26.

**Solution:**

We have,

I =

I =

I =

As we know,

= √cos x|-sin x|

= √cos x|sin x|

= f(x)

So, f(x) is an odd function.

Therefore we get,

I =

I =

As we know, ,

I =

I =

Let cos x = z

^{2}. So, we have=> – sin x dx = 2z dz

Now, the lower limit is, x = 0

=> z

^{2}= cos x=> z

^{2}= cos 0=> z

^{2}= 1=> z = 1

Also, the upper limit is, x = π/2

=> z

^{2}= cos x=> z

^{2}= cos π/2=> z

^{2}= 0=> z = 0

So, the equation becomes,

I =

I =

I =

I =

I =

I = – π/2(log1 – log0) + π/2(log1 – log2) + π(tan

^{-1}0 – tan^{-1}1)I = – π/2[0 – ∞] + π/2(0 – log2) + π(0 – π/4)

I = -∞ – π/2 log2 – π

^{2}/4I = –∞

### Question 27.

**Solution:**

We have,

I =

I =

As we know,

I =

I =

I =

I = 4 – 1

I = 3

### Question 28.

**Solution:**

We have,

I =

I =

As we know, π ≤ x ≤ 2π

=> –2π ≤ –x ≤ –π

=> 0 ≤ 2π – x ≤ π

Therefore, we get

I =

I =

I = 1/2( π – 0) – 1/2(0 – π)

I = π

^{2}/2 + π^{2}/2I = π

^{2}

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