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Worked solution

Two antiderivatives and their constant difference

You are given that I=∫x(1−x)1/2 dxI=\displaystyle\int x(1-x)^{1/2}\,dx.

(i) Use integration by parts to find an expression for II.

(ii) Use the substitution u2=1−xu^2=1-x to find another expression for II.

(iii) Show algebraically that your answers to parts (i) and (ii) differ by a constant.

[N2019/II/1]

CJC · Discussion · Question 11

Answer

(i) −2x3(1−x)3/2−415(1−x)5/2+C1-\dfrac{2x}3(1-x)^{3/2}-\dfrac4{15}(1-x)^{5/2}+C_1.

(ii) −215(1−x)3/2(3x+2)+C2-\dfrac2{15}(1-x)^{3/2}(3x+2)+C_2.

(iii) Difference C1−C2C_1-C_2.

(i)

Choose u=x and dv/dx=(1−x)^(1/2), so du/dx=1 and v=−2(1−x)3/2/3v=-2(1-x)^{3/2}/3. Parts yields I=−2x(1−x)3/2/3−∫[−2(1−x)3/2/3] dxI=-2x(1-x)^{3/2}/3-\int[-2(1-x)^{3/2}/3]\,dx. The remaining integral gives I=−2x3(1−x)3/2−415(1−x)5/2+C1I=-\dfrac{2x}3(1-x)^{3/2}-\dfrac4{15}(1-x)^{5/2}+C_1.

How did this part go?

(ii)

Use the specified substitution u2=1−xu^2=1-x, taking u=1−x≥0u=\sqrt{1-x}\geq0. Then x=1−u2x=1-u^2 and dx/du=−2udx/du=-2u. Therefore I=∫(1−u2)u(−2u) du=∫(2u4−2u2) duI=\int(1-u^2)u(-2u)\,du=\int(2u^4-2u^2)\,du. This gives I=2u5/5−2u3/3+C2=25(1−x)5/2−23(1−x)3/2+C2I=2u^5/5-2u^3/3+C_2=\dfrac25(1-x)^{5/2}-\dfrac23(1-x)^{3/2}+C_2. Factor to obtain I=−215(1−x)3/2(3x+2)+C2I=-\dfrac2{15}(1-x)^{3/2}(3x+2)+C_2.

How did this part go?

(iii)

Factor the nonconstant part of (i): −215(1−x)3/2[5x+2(1−x)]=−215(1−x)3/2(3x+2)-\dfrac2{15}(1-x)^{3/2}[5x+2(1-x)]=-\dfrac2{15}(1-x)^{3/2}(3x+2). Subtracting the factored expression in (ii) cancels it completely, leaving C1−C2C_1-C_2, a constant.

How did this part go?

Common mistakes

The substitution u²=1−x needs a branch when identifying √(1−x) with u; use u≥0. Integration is on intervals with x<1.

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