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  • What is the Integral of 2^(x)? - Physics Forums
    The integral of 2^x can be solved using the relationship between derivatives and integrals The derivative of 2^x is ln(2) * 2^x, which leads to the conclusion that the anti-derivative is (1 ln(2)) * 2^x
  • Can the Volume of a Sphere be Calculated by . . . - Physics Forums
    The the area integral is a double integral over r and the azimuthal angle The volume integral is over r, the azimuthal angle and the polar angle You're missing some parts of the "big picture" about spherical coordinates You kind of skipped the intergrating over the azimuthal angle part by including 2 \pi straight into your integrand
  • What is the Integral of -e^ (-x)? - Physics Forums
    The integral of -e^ (-x) is derived using u-substitution, where u = -x, leading to the result -e^ (-x) + C The negative sign arises because the derivative of -x is -1, thus requiring division by -1 during integration The discussion emphasizes that integration can be viewed as the reverse of differentiation, which simplifies the process Some participants clarify that the method of
  • Integrate exp(-x^2), dx - Physics Forums
    This integral can be done the same way that the integral of exp (-x 2) can be done First, write the integral of x 2 exp (-x 2) from zero to infinity Then write the integral of y 2 exp (-y 2) from zero to infinity (they're both exactly the same as your integral) Now multiply the integrands together double integrate over x and y
  • How do you integrate (r^2-x^2)^ (1 2) with a constant r?
    Jun 18, 2004 #1 FUNKER 120 0 how would you integrate (r^2-x^2)^ (1 2) where r is constant thanks peace Mathematics news on Phys org AI math genius delivers 100% accurate results Robust 'Huber mean' for geometric data protects against noise and outliers Diverse particles form identical geometric patterns when confined, model reveals
  • Finding Volume Under Cone Above Disk - Physics Forums
    Use polar coordinates to find the volume of the given solid: Under the cone z = Sqrt [x^2 + y^2] Above the disk x^2 + y^2 <= 4 2 The attempt at a solution I tried using formatting but I couldn't get it right so I'll explain I changed variables by making the upper and lower limit of the inner integral [-2,2], with the outer integral [0,2pi]





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