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Calculus · real student question

Evaluate the integral of 5 y^(2/5) with respect to y.

Question

Evaluate

∫5y2/5 dy\int 5y^{2/5}\,dy

Step-by-step solution

  1. Pull the constant outside. A numerical factor passes straight through an integral:

    ∫5y2/5 dy=5∫y2/5 dy\int 5y^{2/5}\,dy=5\int y^{2/5}\,dy

  2. Apply the power rule for integration. For n≠−1n\neq -1:

    ∫yn dy=yn+1n+1+C\int y^n\,dy=\frac{y^{n+1}}{n+1}+C

    Here n=25n=\tfrac25, which is safely away from the excluded value −1-1, so the rule applies.

  3. Add one to the exponent. Using a common denominator of 55:

    25+1=25+55=75\frac{2}{5}+1=\frac{2}{5}+\frac{5}{5}=\frac{7}{5}

    ∫y2/5 dy=y7/57/5\int y^{2/5}\,dy=\frac{y^{7/5}}{7/5}

  4. Turn the division by a fraction into a multiplication. Dividing by 75\tfrac75 is multiplying by 57\tfrac57:

    5⋅57y7/5=257y7/55\cdot\frac{5}{7}y^{7/5}=\frac{25}{7}y^{7/5}

    so the antiderivative is 257y7/5+C\tfrac{25}{7}y^{7/5}+C.

  5. Check by differentiating back. The derivative must reproduce the integrand:

    ddy(257y7/5)=257⋅75y2/5=5y2/5✓\frac{d}{dy}\left(\frac{25}{7}y^{7/5}\right)=\frac{25}{7}\cdot\frac{7}{5}y^{2/5}=5y^{2/5}\qquad\checkmark

    Numerically at y=2y=2 the derivative evaluates to 5⋅20.4≈6.59755\cdot 2^{0.4}\approx 6.5975, matching a numerical difference quotient of the antiderivative ✓\checkmark.

Answer

∫5y2/5 dy=257y7/5+C\int 5y^{2/5}\,dy=\frac{25}{7}y^{7/5}+C

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