m almost done and have had problems the entire way through. When I first started this app was awesome it could any simple math up to calculus with no problems. Things like basic integrals were no problem for this app but once you start to do more difficult or complex integral the app ...
integral Given a function f(x)f(x) of a real variable xx and an interval [a,b][a,b] of the real line, the integral ∫baf(x)dx∫abf(x)dx is equal to the area of a region in the xy-plane bounded by the graph of ff, the x-axis, and the vertical lines x=ax=a and x=bx...
tande-3t3. -te-3t3-∫-13e-3tdt −te−3t3−∫−13e−3tdt-te-3t3-∫-13e-3tdt Since−13-13isconstantwith respect tott, move−13-13out of theintegral. −te−3t3−(−13∫e−3tdt)-e-3t3- Simplify. Tap for more steps... ...
The integral of 1u1u with respect to uu is ln(|u|)ln(|u|). arctan(3x)x−16(ln(|u|)+C)arctan(3x)x-16(ln(|u|)+C)Simplify. arctan(3x)x−16ln(|u|)+Carctan(3x)x-16ln(|u|)+CReplace all occurrences of uu with 9x2+19x2+1. arctan(3x)x−16ln(∣∣9x2+1∣...
Step 2 Split the single integral into multiple integrals.Step 3 Use to rewrite as .Step 4 By the Power Rule, the integral of with respect to is .Step 5 Use to rewrite as .Step 6 By the Power Rule, the integral of with respect to is ....
Dividi il singolo integrale in più integrali. ∫x−1dx+∫−4x−2dx∫x-1dx+∫-4x-2dx L'integrale dix−1x-1rispetto axxèln(|x|)ln(|x|). ln(|x|)+C+∫−4x−2dxln(|x|)+C+∫-4x-2dx Poiché−4-4è costante rispetto axx, sposta−4-4fuori dall'integrale. ...
Find the Integral x^5dx Calculus Examples x5dxx5dx By thePower Rule, theintegralofx5x5with respect toxxis16x616x6. 16x6+C16x6+C Enter a problem...
Split the single integral into multiple integrals. ∫dx+∫−sin2(x)dx∫dx+∫-sin2(x)dxApply the constant rule. x+C+∫−sin2(x)dxx+C+∫-sin2(x)dxSince −1-1 is constant with respect to xx, move −1-1 out of the integral. x+C−∫sin2(x)dxx+C-∫sin2(x)dx...
Como 33é constante com relação a xx, mova 33 para fora da integral. x3ex−(3∫ex(x2)dx)x3ex-(3∫ex(x2)dx)Multiplique 33 por −1-1. x3ex−3∫ex(x2)dxx3ex-3∫ex(x2)dxIntegre por partes usando a fórmula ∫udv=uv−∫vdu∫udv=uv-∫vdu, em que u=x2u=x2 e...
, the integral is equal to the area of a region in the xy- plane bounded by the graph of , the x-axis , and the vertical lines and , with areas below the x-axis being subtracted. the x-intercept of a line or curve is the point where it crosses the x-axis , and the y-...