Porisma

Worked solution

Integrate polynomials and linear inner functions

Integrate the following expressions with respect to xx: (a) 2x2+3x+12x^2+3x+1; (b) (5x+4)5(5x+4)^5; (c) 2e2x−12e^{2x-1}; (d) sin⁡(2−3x)\sin(2-3x).

CJC · O Level revision · Example 29

(a) Integrate each polynomial term

Answer

23x3+32x2+x+C\dfrac23x^3+\dfrac32x^2+x+C.

Split the sum and apply the power rule

∫(2x2+3x+1) dx=∫2x2 dx+∫3x dx+∫1 dx=23x3+32x2+x+C\begin{aligned}&\int(2x^2+3x+1)\,dx\\ & =\int2x^2\,dx+\int3x\,dx+\int1\,dx\\ & =\frac23x^3+\frac32x^2+x+C\end{aligned}

The source includes a single arbitrary constant for the resulting antiderivative.

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(b) Compensate for the inner coefficient five

Answer

(5x+4)630+C\dfrac{(5x+4)^6}{30}+C.

Increase the power and divide by both factors

∫(5x+4)5 dx=(5x+4)6(5)(6)+C=(5x+4)630+C\begin{aligned}&\int(5x+4)^5\,dx\\ & =\frac{(5x+4)^6}{(5)(6)}+C\\ & =\frac{(5x+4)^6}{30}+C\end{aligned}

The new power contributes 6 and the linear inner derivative contributes 5.

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(c) Integrate the scaled exponential

Answer

e2x−1+Ce^{2x-1}+C.

Divide by the inner derivative two

∫2e2x−1 dx=2e2x−12+C=e2x−1+C\begin{aligned}&\int2e^{2x-1}\,dx\\ & =\frac{2e^{2x-1}}2+C\\ & =e^{2x-1}+C\end{aligned}

This cancels the printed factor 2 outside the exponential.

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(d) Preserve both negative signs

Answer

13cos⁡(2−3x)+C\dfrac13\cos(2-3x)+C.

Use the sine antiderivative and inner coefficient minus three

∫sin⁡(2−3x) dx=−cos⁡(2−3x)−3+C=13cos⁡(2−3x)+C\begin{aligned}&\int\sin(2-3x)\,dx\\ & =-\frac{\cos(2-3x)}{-3}+C\\ & =\frac13\cos(2-3x)+C\end{aligned}

The final worked part continues onto PDF/printed 28. The antiderivative minus sign and inner coefficient −3 cancel.

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