A-Level Indefinite integration Practice Questions

Free A-Level Indefinite integration practice questions with full step-by-step worked solutions. Covers power rule, integrate x^n, coefficient, sum rule. Practise exam-style problems and check your method.

power ruleintegrate x^ncoefficientsum ruleconstantlinear
A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
Find ∫x3 dx\int x^{3}\,dx.
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Worked solution

  1. Write down the integral

    ∫x3 dx\int x^{3}\,dx

    Integration is the reverse of differentiation, so we look for a function whose derivative is the integrand.

  2. Recall the power rule for integration

    ∫xn dx=xn+1n+1+c(n≠−1)\int x^{n}\,dx=\dfrac{x^{n+1}}{n+1}+c\quad(n\neq-1)

    For every power except −1-1 we add one to the power and divide by the new power.

  3. Integrate term by term

    14x4\frac{1}{4}x^{4}

    Apply the power rule to each term; any number multiplying xx simply carries through unchanged.

  4. Add the constant of integration

    14x4+c\frac{1}{4}x^{4}+c

    Differentiating a constant gives zero, so when we reverse the process we must include an unknown constant cc.

Answer
14x4+c\frac{1}{4}x^{4}+c
Question 2
explain +c markseasy
When we work out ∫6x dx\int 6x\,dx, why must we add a constant cc?
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Worked solution

  1. Integrate 6x6x

    ∫6x dx=3x2+c\int 6x\,dx=3x^{2}+c

    Add one to the power and divide by 22.

  2. Think about differentiating

    ddx(3x2+7)=6x\frac{d}{dx}(3x^{2}+7)=6x

    Any constant added on differentiates to zero.

  3. Conclude

    3x2+c3x^{2}+c

    Because the constant disappears when differentiating, we cannot know it, so we write +c+c.

Answer
3x2+c3x^{2}+c
Question 3
no product rule for integrals marksintermediate
A student writes ∫(2x−1)(x+3) dx=∫(2x−1) dx×∫(x+3) dx\int (2x-1)(x+3)\,dx = \int(2x-1)\,dx \times \int(x+3)\,dx. Which statement is correct?
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Worked solution

  1. Test the claimed rule

    ∫(2x−1)(x+3) dx\int(2x-1)(x+3)\,dx

    There is no product rule for integration, so the claimed shortcut is suspect.

  2. Expand the brackets

    2x2+6x−x−32x^{2}+6x-x-3

    Multiply each term of the first bracket by each term of the second.

  3. Collect like terms

    2x2+5x−32x^{2}+5x-3

    Combine 6x−x=5x6x-x=5x to get a tidy quadratic.

  4. Recall the power rule

    ∫xn dx=xn+1n+1+c(n≠−1)\int x^{n}\,dx=\dfrac{x^{n+1}}{n+1}+c\quad(n\neq-1)

    Every term is now a power of xx.

  5. Integrate term by term

    2x33+5x22−3x\frac{2x^{3}}{3}+\frac{5x^{2}}{2}-3x

    Raise each power by one and divide by the new power.

  6. Add the constant

    2x33+5x22−3x+c\frac{2x^{3}}{3}+\frac{5x^{2}}{2}-3x+c

    The correct method expands first, so the student's product shortcut is wrong.

Answer
2x33+5x22−3x+c\frac{2x^{3}}{3}+\frac{5x^{2}}{2}-3x+c
Question 4
interpret constant of integration markshard
Two curves have the same gradient function dydx=3x2−2\dfrac{dy}{dx}=3x^{2}-2. What is the geometric relationship between them?
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Worked solution

  1. Integrate the gradient

    y=∫(3x2−2) dxy=\int(3x^{2}-2)\,dx

    Reversing differentiation recovers yy from its gradient function.

  2. Apply the power rule

    ∫3x2 dx=x3\int 3x^{2}\,dx=x^{3}

    Raise the power to 33 and divide by 33.

  3. Integrate the constant term

    ∫−2 dx=−2x\int -2\,dx=-2x

    A constant integrates to the constant times xx.

  4. Write the general solution

    y=x3−2x+cy=x^{3}-2x+c

    Every curve with this gradient has this form; only cc can differ.

  5. Take a first curve

    y1=x3−2x+c1y_{1}=x^{3}-2x+c_{1}

    Choose one particular value of the constant.

  6. Take a second curve

    y2=x3−2x+c2y_{2}=x^{3}-2x+c_{2}

    Choose a different value of the constant.

  7. Subtract the two curves

    y1−y2=c1−c2y_{1}-y_{2}=c_{1}-c_{2}

    The variable parts cancel, leaving a constant difference.

  8. Interpret the constant difference

    y1=y2+(c1−c2)y_{1}=y_{2}+(c_{1}-c_{2})

    Adding a constant to every yy-value shifts the whole curve up or down.

  9. Interpret geometrically

    vertical translation\text{vertical translation}

    A fixed difference in height at every xx is exactly a vertical shift.

  10. Conclude

    y=x3−2x+cy=x^{3}-2x+c

    So the curves are vertical translations of one another.

Answer
y=x3−2x+cy=x^{3}-2x+c
Question 5
8 markschallenging
A curve has dydx=5xx−6x+3\dfrac{dy}{dx}=5x\sqrt{x}-6\sqrt{x}+3 and passes through (1,4)(1,4). Find its equation and yy when x=4x=4.
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Worked solution

  1. Write down the gradient function

    dydx=5xx−6x+3\frac{dy}{dx}=5x\sqrt{x}-6\sqrt{x}+3

    The derivative gives the gradient of the curve; integrating it reverses differentiation to recover yy.

  2. Write the surd terms as powers

    dydx=5x32−6x12+3\frac{dy}{dx}=5x^{\frac{3}{2}} - 6x^{\frac{1}{2}} + 3

    Use xx=x3/2x\sqrt{x}=x^{3/2} and x=x1/2\sqrt{x}=x^{1/2}.

  3. Integrate to find yy (remember +c+c)

    y=∫(5x32−6x12+3) dxy=\int\left(5x^{\frac{3}{2}} - 6x^{\frac{1}{2}} + 3\right)\,dx

    Reversing differentiation always introduces an unknown constant, so we carry a +c+c until a point lets us find it.

  4. Recall the power rule

    ∫xn dx=xn+1n+1+c(n≠−1)\int x^{n}\,dx=\dfrac{x^{n+1}}{n+1}+c\quad(n\neq-1)

    Add one to each power and divide by the new power.

  5. Integrate this term

    ∫5x32 dx=2x52\int 5x^{\frac{3}{2}}\,dx = 2x^{\frac{5}{2}}

    Add one to the fractional power and divide by the new power; index laws work exactly as they do for whole-number powers.

  6. Integrate this term

    ∫−6x12 dx=−4x32\int -6x^{\frac{1}{2}}\,dx = -4x^{\frac{3}{2}}

    Add one to the fractional power and divide by the new power; index laws work exactly as they do for whole-number powers.

  7. Integrate this term

    ∫3 dx=3x\int 3\,dx = 3x

    A constant integrates to that constant multiplied by xx.

  8. Write the general solution

    y=2x52−4x32+3x+cy=2x^{\frac{5}{2}} - 4x^{\frac{3}{2}} + 3x+c

    Every curve in this family has the given gradient function; the constant cc decides which one we have.

  9. Use the point on the curve

    4=1+c4=1+c

    The curve passes through (1, 4)(1,\,4), so substituting x=1x=1 must give y=4y=4.

  10. Solve for the constant

    c=3c=3

    Rearranging fixes the single constant that makes the curve pass through the given point.

  11. Substitute the constant back

    y=2x52−4x32+3x+3y=2x^{\frac{5}{2}} - 4x^{\frac{3}{2}} + 3x+3

    Replace cc with the value we found to get the specific curve, not the whole family.

  12. State the equation of the curve

    y=2x52−4x32+3x+3y=2x^{\frac{5}{2}} - 4x^{\frac{3}{2}} + 3x + 3

    Substituting cc back gives the one curve with this gradient that passes through the point.

  13. Check by differentiating

    dydx=5x32−6x12+3\frac{dy}{dx}=5x^{\frac{3}{2}} - 6x^{\frac{1}{2}} + 3

    Differentiating our equation returns the original gradient function, confirming the working.

  14. Confirm the point lies on the curve

    x=1:  y=4x=1:\; y=4

    Substituting the xx-coordinate reproduces the given yy-coordinate, as required.

  15. Evaluate yy at x=4x=4

    x=4:  y=47x=4:\; y=47

    Substitute the required xx-value into the equation of the curve.

  16. State both required results

    y=2x52−4x32+3x+3andy(4)=47y=2x^{\frac{5}{2}} - 4x^{\frac{3}{2}} + 3x + 3\quad\text{and}\quad y(4)=47

    The question asked for the equation of the curve and the value of yy at the given point; here are both.

Answer
y=2x52−4x32+3x+3y=2x^{\frac{5}{2}} - 4x^{\frac{3}{2}} + 3x + 3

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