integral from 0 to a Power[\(40)-Divide[Power[a,2],2]+Divide[a,2]x\(41),2] dx

asked by guest
on Nov 16, 2024 at 5:50 am



You asked:

Evaluate the integral: ∫0a(−(a22)+a2x)21dx\int_{0}^{a {\left( -\left( \frac{{a}^{2}}{2} \right) + \frac{a}{2} x \right)}^{2}}{1}dx

MathBot Answer:

Evaluated



∫0a(−(a22)+a2x)21dx=a(−a22+ax2)2\displaystyle \int_{0}^{a {\left( -\left( \frac{{a}^{2}}{2} \right) + \frac{a}{2} x \right)}^{2}}{1}dx = a \left(- \frac{a^{2}}{2} + \frac{a x}{2}\right)^{2}

Expanded

∫0a(−(a22)+a2x)21dx=∫0a54−a4x2+a3x241 dx\int_{0}^{a {\left( -\left( \frac{{a}^{2}}{2} \right) + \frac{a}{2} x \right)}^{2}}{1}dx = \int\limits_{0}^{\frac{a^{5}}{4} - \frac{a^{4} x}{2} + \frac{a^{3} x^{2}}{4}} 1\, dx

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