Diff Eq: Solve y'+(1/t)y=3cos(2t)?
Multiply both sides by the integrating factor exp(∫ dt/t) = exp(ln t) = t: ty' + y = 3t cos(2t) ==> (d/dt)(ty) = 3t cos(2t), by the product rule Integrate both sides: ty = ∫ 3t cos(2t) dt. We evaluate the right side with integration by parts. Let u = 3t, dv = cos(2t) dt du = 3 dt, v = (1/2) sin(2t) ∫ 3t cos(2t) dt = (3/2) t sin(2t) - ∫(3/2) sin(2t) dt = (3/2) t sin(2t) + (3/4) cos(2t) + C Thus, the general solution is ty = (3/2) t sin(2t) + (3/4) cos(2t) + C ==> y = (3/2) sin(2t) + (3/4) cos(2t)/t + C/t. I hope this helps!
Evaluate the integral?
Factor out constants 3 * Integral ( t * sin(5t) ) Let u = t, du = dt, dv = sin(5t) dt, v = -0.2 * cos(5t) Use the formula: uv - Integral (v du) 3 * ( -0.2t * cos(5t) ) - Integral ( -0.2 * cos(5t) ) Integrate 3 * ( -0.2t * cos(5t) ) + ( 0.04 * sin(5t) ) Plug in pi and 0 3 * ( -0.2pi * cos(5pi) ) + ( .04 * sin(5pi) ) - ( -0.2(0) * cos(5 * 0) ) + ( .04 * sin(5 * 0) ) Evaluate 3 * ( -0.2pi * -1 ) + ( .04 * 0 ) - ( 0 * -1 ) + ( .04 * 0 ) Multiply ( 0.6 pi ) + ( 0 ) + ( 0 ) + ( 0 ) Add 0.6pi I hope this helps. Have a good day.
How do I integrate [math]e^{x^3}[/math]?
Let [math]I=\int e^{x^{3}}dx[/math]Take [math]t=x^3[/math][math]\therefore t^{\frac{1}{3}}=x[/math]Differentiating [math]t=x^3 [/math]w.r.t. [math]x[/math], we get,[math]\frac{dt}{dx}=3x^2[/math][math]dx=\frac{dt}{3x^2}[/math][math]dx=\frac{dt}{3t^{\frac{2}{3}}}[/math][math]\therefore I=\int e^{t}t^{-\frac{2}{3}}dt[/math]If you try to integrate it using the UV rule, it will not end. This is a special type of integral whose answer is [math]\frac{\gamma (\frac{1}{3} ,-x^3)}{3}+c[/math]Hope this helps! You can also get such questions solved and get the detailed solution within seconds using the Scholr app by just uploading a picture of the question and also get to be a part of an ever-growing community of students.
How do I solve this integration [math] 4t^2 - 3t + 2 [/math] ,with the limit 1 to 3?
Int (4t^2 -3t+2 )dt |1to 3|= (4t^3/3–3t^2/2+2t ) |1to3|=[4(3)^3/3–3(3)^2/2+2(3)]-[4(1)^3/3–3(1)^2/2+2(1)]=(36–27/2+6)-(4/3–3/2+2)= (42–27/2)-(11/6)=(252–92) /6=160/6=26.66
Find the interval on which the curve y(x) is concave upward. (Enter INFINITY for and -INFINITY for -.) ?
y(x) is a function that is defined by an integral: y(x) = integral from 0 to x of 1/(3+t+3t^2) dt By the Fundamental Thm of Calculus, the derivative of this function is: y '(x) = 1/( 3 + x + 3x^2 ) A function is concave up when the second derivative is positive, so we need the second derivative: y ''(x) = ( -1 ) *[ 1/ ( 3+x+3x^2)^2 ] * ( 1 + 6x ) = {chain rule} -(1+6x) * [1/ ( 3+x+3x^2)^2 ] Now the second term is a square and so is always nonnegative. Therefore, the second derivative will be positive when: - (1 + 6x) > 0 or 1+6x < 0 which means x < -1/6 So it's concave up when x < -1/6
Evaluate the integrals?
1.) integral 1 to 4 : (5-2t + 3t^2) dt = 5(t/1) - 2(t^2 / 2) + 3(t^3 / 3) evaluated from 1 to 4 = 5t - t^2 + t^3 evaluated from 1 to 4 = [5(4) - 4^2 + 4^3] - [5(1) - 1^2 +1^3] = [20 - 16 + 64] - [5 - 1 +1] = 68 - 5 = 63 2.) integral 0 to 1 : x^4/5 dx = x^(4/5 + 1) / (4/5 + 1) evaluated from 0 to 1 = 5(x^9/5) / 9 evaluated from 0 to 1 = 5(1^9/5) / 9 - 5(0^9/5) / 9 = 5/9 - 0 = 5/9 3.) integral 1 to 2 : 3/t^4 dt = integral 1 to 2 : 3t^-4 dt = 3t^-3 / -3 evaluated from 1 to 2 = -t^-3 evaluated from 1 to 2 = -1/t^3 evaluated from 1 to 2 = [-1/(2)^3] - [-1/(1)^3] = [-1/8] - [-1] = 1 - 1/8 = 7/8 I hope I got it right. It's really difficult to evaluate integrals using a text editor.
Help in definite integral: t^3 ln(2 t) dt over interval (7,1)?
Use integration by parts. u=ln(2t) du=1/t dv= t^3 v=.25t^4 .25t^4ln(2t)-(1/16)t^4 plug in x limits of integration solution: [(7^4)ln(14)]/4-(7^4)/16-(ln2)/4+ 1/16 this approximately equals 1433.921
How do I integrate[math] \dfrac{x^3}{x^2-2x+1}[/math]?
Let's add and subtract terms in numerator to obtain simplest integrals.[math] \displaystyle\int \dfrac{x^3}{x^2 - 2x + 1} dx [/math][math]= \displaystyle\int \dfrac{x^3 - 2x^2 + x + 2x^2 - x}{x^2 - 2x + 1} dx [/math][math]= \displaystyle\int \dfrac{2x^2 - x}{x^2 - 2x + 1} dx + \displaystyle\int x dx [/math][math]= \displaystyle\int \dfrac{2x^2 - 4x + 2 + 3x - 2}{x^2 - 2x + 1} dx + \displaystyle\int x dx [/math][math]= \displaystyle\int \dfrac{3x - 2}{x^2 - 2x + 1} dx + \displaystyle\int (x + 2) dx [/math][math]= \displaystyle\int \dfrac{3x - 3 + 1}{x^2 - 2x + 1} dx + \displaystyle\int (x + 2) dx [/math][math]= \displaystyle\int \dfrac{3}{x - 1} dx + \displaystyle\int\dfrac{1}{(x - 1)^2} dx + \displaystyle\int (x + 2) dx [/math][math]= 3 \ln|x - 1| - \dfrac{1}{x - 1} + \dfrac{x^2}{2} + 2x + C[/math]Hope you find this helpful :-)
Find the surface area generated by rotating the given curve about the y-axis.?
From ∫(0 to 5) 36π t^3 √(1 + t^2) dt, rewrite this as ∫(0 to 5) 18π t^2 (2t √(1 + t^2) dt) Use integration by parts: u = 18πt^2, dv = 2t (1 + t^2)^(1/2) dt du = 36πt dt, v = (2/3)(1 + t^2)^(3/2) ==> 12πt^2 (1 + t^2)^(3/2) {for t = 0 to 5} - ∫(0 to 5) 24π t(1 + t^2)^(3/2) dt = 12πt^2 (1 + t^2)^(3/2) - 12π * (2/5)(1 + t^2)^(5/2) {for t = 0 to 5} = (12π/5) (1 + t^2)^(3/2) [5t^2 - 2(1 + t^2)] {for t = 0 to 5} = (12π/5) (1 + t^2)^(3/2) (3t^2 - 2) {for t = 0 to 5} = (12π/5) [73 * 26^(3/2) + 2]. I hope this helps!
The displacement of a particle is given by x=(t-2) ^2. What is the distance covered by the particle in the first 4 seconds?
Although you can refer the answer given by Mr Kulkarni but if you want to go logically then just try to imagine the graph of x versus time of the displacement of this object initially the object was 4 unit away from origin but after 2 second it comes to origin and then again after 2 seconds it gets back to its original coordinates hence the total distance travelled is equals to 4 + 4 that is 8 units but if you want to know that how will you do a complicated version of this problem that is in which the XT function is not such simple then either as I told you should refer Mr Kulkarni’s method and you also have one more approach you just need to find the region of the curve where the velocity is negative that is slope of the XT curve is negative which is in between the region of zero to 2 seconds in the given problem you just break the problem in two regions from initial stage to the stage where the slope becomes zero and then again from the time wear 2.0 to the final time if the slope becomes zero more time in between the duration then you can repeat this again and again and break the problem in as many parts as slope become zero in the duration.Hope it helped.