Hard A-Level Indefinite integration Questions

Write down the gradient function

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

Write the surd terms as powers

Use $x\sqrt{x}=x^{3/2}$ and $\sqrt{x}=x^{1/2}$.

Integrate to find $y$ (remember $+c$)

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

Recall the power rule

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

Integrate this term

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

Integrate this term

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

Integrate this term

A constant integrates to that constant multiplied by $x$.

Write the general solution

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

Use the point on the curve

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

Solve for the constant

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

Substitute the constant back

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

State the equation of the curve

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

Check by differentiating

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

Confirm the point lies on the curve

Substituting the $x$-coordinate reproduces the given $y$-coordinate, as required.

Evaluate $y$ at $x=4$

Substitute the required $x$-value into the equation of the curve.

State both required results

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

State the general principle

The rule needs the denominator $n+1$ to be non-zero.

Identify the danger

If $n=-1$ the denominator becomes zero, so the rule breaks down.

Look at the awkward case

This is $\int \tfrac{1}{x}\,dx$, which the power rule cannot handle.

Try the (invalid) rule anyway

Dividing by zero is undefined, so the power rule gives nonsense here.

Test other negative powers

For $n=-2$ the rule works fine because $n+1=-1\neq 0$.

Test a fractional power

Fractional powers are fine as long as $n\neq -1$.

Summarise which powers are safe

Every real power except $-1$ can be integrated with the power rule.

Explain why $-1$ is special

The antiderivative of $x^{-1}$ is a logarithm, met later in the course, not a power of $x$.

Confirm with differentiation

Differentiating the rule's output returns $x^{n}$ whenever $n+1\neq 0$.

Re-examine $n=-1$

The check itself fails at $n=-1$, confirming the exclusion.

Consider a student's slip

A student who ignores the restriction reaches an undefined expression.

Correct the misconception

The restriction is essential, not optional.

Generalise

A constant multiplier simply carries through the same rule.

Note the single exception again

Only this one value is forbidden.

Conclude

The condition $n\neq -1$ is exactly what keeps the power rule valid.

Write down the integral

There is no product rule for integration, so we expand first.

Multiply the brackets

Use each term of the first bracket on each term of the second.

Collect like terms

Combine the two middle terms $2x-3x=-x$.

Recall the power rule

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

Integrate the $x^{2}$ term

Raise the power to $3$ and divide by $3$.

Integrate the $x$ term

Raise the power to $2$ and divide by $2$.

Integrate the constant

A constant integrates to the constant times $x$.

Combine the integrated terms

Add the three integrated pieces together.

Add the constant of integration

Include $+c$ for the indefinite integral.

Begin the check by differentiating

Differentiate the first term to test our answer.

Differentiate the second term

This should match the middle term of the integrand.

Differentiate the third term

This matches the constant term.

Differentiate the constant

The constant disappears, as expected.

Reassemble the derivative

This equals the expanded integrand exactly.

Conclude

Since differentiating returns the integrand, the result is proven correct.

Write down the second derivative

The second derivative is the rate of change of the gradient; integrating once gives the gradient function.

Recall the power rule

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

Integrate this term (first integration)

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

Write the gradient with its constant

The first integration introduces its own constant, which we call $A$.

Use the first condition

At $x=1$ the gradient is $2$, which lets us find $A$.

Solve for $A$

This pins down the gradient function completely.

State the gradient function

Now we know the exact gradient function, ready to integrate a second time.

Integrate again to find $y$ (add $B$)

Integrating the gradient recovers $y$, introducing a second constant $B$.

Integrate this term (second integration)

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

Integrate this term (second integration)

A constant integrates to that constant multiplied by $x$.

Write $y$ with its constant

This is the general form of $y$; the second condition will fix $B$.

Use the second condition

At $x=1$ we know $y=1$, which determines $B$.

Solve for $B$

Both constants are now known.

State the equation of the curve

Putting $A$ and $B$ back gives the single curve satisfying every condition.

Check by differentiating twice

Differentiating $y$ twice returns the original second derivative, confirming the result.

Write down the second derivative

The second derivative is the rate of change of the gradient; integrating once gives the gradient function.

Recall the power rule

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

Integrate this term (first integration)

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

Integrate this term (first integration)

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

Write the gradient with its constant

The first integration introduces its own constant, which we call $A$.

Use the first condition

At $x=1$ the gradient is $0$, which lets us find $A$.

Solve for $A$

This pins down the gradient function completely.

State the gradient function

Now we know the exact gradient function, ready to integrate a second time.

Integrate again to find $y$ (add $B$)

Integrating the gradient recovers $y$, introducing a second constant $B$.

Integrate this term (second integration)

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

Integrate this term (second integration)

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

Integrate this term (second integration)

A constant integrates to that constant multiplied by $x$.

Write $y$ with its constant

This is the general form of $y$; the second condition will fix $B$.

Use the second condition

At $x=0$ we know $y=2$, which determines $B$.

Solve for $B$

Both constants are now known.

State the equation of the curve

Putting $A$ and $B$ back gives the single curve satisfying every condition.

Check by differentiating twice

Differentiating $y$ twice returns the original second derivative, confirming the result.

Evaluate $y$ at $x=2$

Substitute the required $x$-value into the equation of the curve.