GSEB Solutions Class 12 Maths Chapter 7 Integrals Ex 7.1

Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 7 Integrals Ex 7.1 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 11 Maths Chapter 7 Integrals Ex 7.1

Find an antiderivative (or integral) of the following by the method of inspection:
Question 1.
sin 2x
Solution:
We know that $\frac{d}{dx}$ cos 2x = – 2sin 2x.
or $\frac{d}{dx}$ (- $\frac{1}{2}$ cos 2x) = sin 2x
∴ ∫sin 2x dx = – $\frac{1}{2}$ cos 2x + C.

Question 2.
cos 3x
Solution:
2. We know that $\frac{d}{dx}$ (sin 3x) = 3 cos 3x.
⇒ cos 3x = $\frac{1}{3}$ $\frac{d}{dx}$ (sin 3x)
⇒ cos 3x = $\frac{d}{dx}$ ($\frac{1}{3}$ sin 3x)
∴ An antiderivative of cos 3x is $\frac{1}{3}$ sin 3x + C.

By using free Taylor Series Calculator, you can easily find the approximate value of the integration function

Question 3.
e^2x
Solution:
We know that
$\frac{d}{dx}$(e^2x) = 2e^2x.
⇒ e^2x = $\frac{1}{2}$ $\frac{d}{dx}$ (e^2x)
⇒ e^2x = $\frac{d}{dx}$ ($\frac{1}{2}$ e^2x)
∴ An antiderivative of e^2x is $\frac{1}{2}$ e^2x + C.

Question 4.
(ax + b)²
Solution:
We know that $\frac{d}{dx}$ (ax + b)³ = 3a(ax + b)².
⇒ (ax + b)² = $\frac{1}{3a}$ $\frac{d}{dx}$(ax + b)³
⇒ (ax + b)² = $\frac{d}{dx}$[$\frac{1}{3a}$(ax + b)³]
∴ An antiderivative of (ax + b)² = $\frac{1}{3a}$(ax + b)³ + C.

Question 5.
sin 2x – 4e^3x
Solution:
We know that $\frac{d}{dx}$(cos 2x) = – 2 sin 2x.
⇒ sin 2x = $\frac{d}{dx}$(-$\frac{1}{2}$ cos 2x)
and $\frac{d}{dx}$(4e^3x) = 4 × 3e^3x
⇒ 4e^3x = $\frac{d}{dx}$($\frac{1}{3}$ e^3x)
∴ An antiderivative of sin 2x – 4e^3x is – $\frac{1}{2}$ cos 2x – $\frac{4}{3}$e^3x + C.

Find the following integrals:
Question
6. ∫(4e^3x + 1)dx
Solution:

Question 7.
∫x²(1 – $\frac{1}{x^{2}}$)dx
Solution:

Question 8.
∫(ax² + bx + c)dx
Solution:

Question 9.
∫(2x² + e^x) dx
Solution:

Question 10.
∫($\sqrt{x}$ – $\frac{1}{\sqrt{x}}$)² dx
Solution:

Question 11.
∫$\frac{x^{3}+5 x^{2}-4}{x^{2}}$dx
Solution:

Question 12.
∫$\frac{x^{3}+3 x+4}{\sqrt{x}}$dx
Solution:

Question 13.
∫$\frac{x^{3}-x^{2}+x-1}{x-1}$dx
Solution:

Question 14.
∫(1 – x)$\sqrt{x}$ dx
Solution:

Question 15.
∫$\sqrt{x}$(3x² + 2x + 3)dx
Solution:

Question 16.
∫(2x – 3cosx + e^x)dx
Solution:

Question 17.
∫(2x² – 3sinx + 5$\sqrt{x}$)dx
Solution:

Question 18.
∫secx(sec x + tan x)dx
Solution:

Question 19.
∫$\frac{sec^{2}x}{cosec^{2}x}$dx.
Solution:

Question 20.
∫$\frac{2-3 \sin x}{\cos ^{2} x}$ dx.
Solution:

Choose the correct answers in the following questions 21 and 22:
Question 21.
The antiderivative of ($\sqrt{x}$ + $\frac{1}{\sqrt{x}}$) equals
(A) $\frac{1}{3}$x^1/3 + 2x^1/2 + C
(B) $\frac{2}{3}$x^2/3 + $\frac{1}{2}$x² + C
(C) $\frac{2}{3}$x^3/2 + 2x^1/2 + C
(D) $\frac{3}{2}$x^3/2 + $\frac{1}{2}$x^1/2 + C
Solution:

⇒ Part(C) is the correct answer.

Question 22.
If $\frac{d}{dx}$ f(x) = 4x³ – $\frac{3}{x^{4}}$ such that f(2) = 0, then f(x) is
(A) x⁴ + $\frac{1}{x^{3}}$ – $\frac{129}{8}$
(B) x³ + $\frac{1}{x^{4}}$ + $\frac{129}{8}$
(C) x⁴ + $\frac{1}{x^{3}}$ + $\frac{129}{8}$
(D) x³ + $\frac{1}{x^{4}}$ – $\frac{129}{8}$
Solution:

Putting C = – $\frac{129}{8}$ in (1), we get
f(x) = x⁴ + $\frac{1}{x^{3}}$ – $\frac{129}{8}$
⇒ Part(A) is the correct answer.