\[\int_{-2}^1 5 d x \text { will be }\] ....

Channel:
PW Solutions
Subscribers:
445,000
Published on ● Video Link: https://www.youtube.com/watch?v=sq3hfkHEI7U


Duration: 1:32
0 views
0

\[\int_{-2}^1 5 d x \text { will be }\]  📲PW App Link - https://bit.ly/YTAI_PWAP 🌐PW Website - https://www.pw.live



Other Videos By PW Solutions


2024-05-09Find \(\frac{d z}{d x}\) at \((x, y, z)=(1,0,1)\) for given equation \(x^2 \cos (y)=\sin \left(y....
2024-05-09Joule second is the unit of....
2024-05-09Evaluate \(\int_0^{\pi / 6}[2 \cos (2 x)-3 \sin (3 x)] d x\)....
2024-05-09A particle moves along \(x\)-axis with acceleration \(a=6(t-1)\), where \(t\) is in seconds. If ....
2024-05-09Current is flowing in conductor as \(i=6 t+9 t^2\) here \(\mathrm{t}\) is time and \(\mathrm{i}\....
2024-05-09Find \(\left|\frac{d y}{d x}\right|\) at \((0,3)\) for given equation \(y^2 e^{2 x}=3 y+x^2\)....
2024-05-09A body is moving along \(x\) axis as \(v=2 t+3 t^2+2\) here \(v\) is velocity and \(t\) is time ....
2024-05-09\[\int_{-2}^{+2}\left(t^2-1\right) d t\] ....
2024-05-09Use a definite integral to find the area of the region between the given curve and the \(x\)-axi....
2024-05-09\(\int_{-4}^{-1} \frac{\pi}{2} d \theta\) will be ....
2024-05-09\[\int_{-2}^1 5 d x \text { will be }\] ....
2024-05-09\(\int\left(\sqrt[3]{x}+\frac{1}{\sqrt[3]{x}}\right) d x\) will be ....
2024-05-09\(\int(3 \sin x) d x\) will be ....
2024-05-09\(\int\left(\frac{1}{3 x}\right) d x\) will be ....
2024-05-09\(\int\left(-\frac{1}{2} x^{-3 / 2}\right) d x\) will be ....
2024-05-09\(\int\left(\frac{1}{3 \sqrt[3]{x}}\right) d x\) will be ....
2024-05-09\(\int\left(\frac{1}{2} x^{-1 / 2}\right) d x\) will be ....
2024-05-09Determine the position where potential energy will be minimum if \(U(x)=100-50 x+1000 x^2 \mathr....
2024-05-09\(\int\left(\frac{4}{3} \sqrt[3]{x}\right) d x\) will be ....
2024-05-09\(\int\left(\frac{3}{2 \sqrt{x}}\right) d x\) will be ....
2024-05-09Particle's position as a function of time is given by \(x=-t^2\) \(+4 t+4\) find the maximum val....