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                   indefinite integrals
                   

3 years ago

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Hi Anam,


Please ask what you want to ask so that we can reply to your questions.


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3 years ago

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5 ∫ 0 (x + 1) dx
 
 
integration of x+1 from 0 t0 5 is equal to ((x^2)/2)+x) from 0 to 5 =((5^2)/2+5)-0 =((25/2)+5) =35/2 =17.5
 
Miryala Gopalakrishna 8 months ago
 
Ans:- 5 ∫ 0 (x+1)dx =[(x 2 /2)+x] 0 5 =(25/2)+5 =(25+10)/2 =35/2
 
Anusha 8 months ago
3 ∫ 2 (2x 2 + 1) dx
 
 
integration of (2(X^2)+1)dx from 2 to 3 is equal to (2((x^3)/3)+x) from 2 to3 =(2((3^3)/3)+3)-(2((2^3)/3)+2) =(2*(27/3)+3)-(2*(8/3)+2) =(18+3)-((16/3)+2) =(21)-(22/3) =41/3
 
Miryala Gopalakrishna 8 months ago
 
Ans:- 3 ∫ 2 (2x 2 +1)dx =2[(x 2+1 /2+1)] 2 3 +[x] 2 3 =2[(x 3 /3)] 2 3 +[x] 2 3 =2[(3 3 /3)+(2 3 /3)]+[3-2] =2[(27+8)/3]+1 =2[(35/3)]+1 =(70/3)+1 =73/3
 
Anusha 8 months ago
what is the antiderivative of sinx
 
 
anti derivative means integration and its value equals theta1 theta2 ∫sinx.dx = -cos(theta1-theta2) +c approve if useful
 
grenade one month ago
 
-COSX
 
noogler one month ago
what is delta in conditions for pair of lines?
 
 
delta=0 h 2 -ab>0 pair of intersecting at lines h 2 -ab delta is not eq to 0 a=b , h=0 circle h 2 =ab parabola h 2 -ab h 2 -ab>0 hyperbola approve if useful
 
ng29 one month ago
 
delta=abc+2fgh-af 2 -bg 2 -ch 2 =0 for pair f st.lines delta is not zero for parabola,hyperbola,circle,ellipse
 
noogler one month ago
 
oh really tell me which are those points that are left????????????????????????????????????/
 
ng29 one month ago
limit as x approaches 0 of (e x -1-x)/x 2 ? Do without L Hospital rule.
 
 
in these type of questions u can proceed with series expansion formula e x =1 + x + x 2 /fact2 -------- so on so we get the limit , (x 2 /fact 2--------------)/x 2 so answer is ½ as limit x ...
 
ng29 one month ago
 
take commons and use squaring and finally use the basic prpts of limits and u will get the ans 0.5
 
grenade one month ago
Find the greatest term in the expansion of (7-5x)^2 where x=2/3 Please explain the steps. Thankyou in advance.
 
 
Hello Student. Please find the solution (7-5x)^2 = 49 + 25x^2 – 70x so first term = 49 second term = -70*2/3 third term = 25 x^2 = 25*4/9 = 11.11 So 49 is greatest...
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Nishant Vora 9 months ago
 
Sorry sir.. i am extremely sorry.. Actually i typed in the wrong question.. it was (7-5x)^11, where x=2/3.. I am extremely sorry once again. Sir please give me the correct answer for this.
 
Swagat Bordoloi 9 months ago
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