Mathematics

Evaluate:
$$\displaystyle \int _0^{3} x^2+2x dx$$


SOLUTION
$$\displaystyle \int _0^3 x^2+2x dx$$

By using $$\displaystyle\int{{x}^{n}dx}=\dfrac{{x}^{n+1}}{n+1}+c$$

$$=\left. \dfrac{x^3}3+x^2 \right|_0^3$$

$$= 9+9-0-0=18$$
View Full Answer

Its FREE, you're just one step away


Subjective Medium Published on 17th 09, 2020
Next Question
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86
Enroll Now For FREE

Realted Questions

Q1 Subjective Medium
Evaluate:$$\displaystyle\int_{0}^{\frac{\pi}{2}}{{\sin}^{2}{x}\,dx}$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q2 Subjective Medium
Find an anti derivative (or integral) of the given function by the method of inspection.
$$\cos 3x$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q3 Single Correct Medium
$$\int _{ 0 }^{ a }{ \left( f(x)+f(-x) \right)  } dx=$$
  • A. $$-\int _{ -a }^{ a }{ f(x)dx } $$
  • B. $$0$$
  • C. $$-\int _{ -a }^{ a }{ f(-x)dxz } $$
  • D. $$2\int _{ 0 }^{ a }{ f(x)dx } $$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q4 Single Correct Hard
If $$\displaystyle I = \int \frac {dx}{(2 \sin x + \sec x)^4}$$, then I equals
  • A. $$\displaystyle \frac {1}{5 \tan^5 x} + \frac {1}{3 \tan^6 x} - \frac {I}{(2 \sin x + \sec x)^3} + C$$
  • B. $$\displaystyle \frac {-1}{3(2 \sin x + \sec)^3} + \tan^{-1} (3\sqrt {\tan x}) + C$$
  • C. $$\displaystyle \frac {-1}{3(2 \sin x + \sec x)^3} - \tan^{-1} (3\sqrt {\tan x}) + C$$
  • D. $$\displaystyle -\frac {1}{5 \tan^5 x} + \frac {1}{3 \tan^6 x} - \frac {2}{7 \tan^7 x} + C$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q5 Single Correct Medium
$$\displaystyle \int_2^3\dfrac{dx}{x^2-1}$$
  • A. $$log\dfrac{3}{2}$$
  • B. $$2log\dfrac{3}{2}$$
  • C. $$\log\dfrac{3}{2}$$
  • D. $$\dfrac{1}{2}log\dfrac{3}{2}$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer