Standard integrals
Integration
- AS
- Non-calculator
- 8 questions
- 37 marks
These are original practice questions for the Edexcel A-level Pure content, not a past paper. This site is independent and not affiliated with Pearson.
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Find $\displaystyle \int (6x^2 - 4x + 3)\,dx$.
3 marks
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Find $\displaystyle \int e^{3x}\,dx$ and $\displaystyle \int \cos 2x\,dx$.
3 marks
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Evaluate $\displaystyle \int_0^2 (x^2 + 1)\,dx$.
4 marks
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A curve has gradient $\dfrac{dy}{dx} = 6x^2$ and passes through $(1, 5)$. Find the equation of the curve.
4 marks
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Find $\displaystyle \int \big(4\sin 2x - e^{-x}\big)\,dx$.
5 marks
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The gradient of a curve is $3x^2 - 4$, and the curve passes through $(2, 5)$. Find the equation of the curve.
5 marks
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Find $\displaystyle \int \big(5e^{2x} - 6x^2\big)\,dx$, and hence evaluate $\displaystyle \int_0^1 \big(5e^{2x} - 6x^2\big)\,dx$.
6 marks
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A curve has $\dfrac{dy}{dx} = 3x^2 - \dfrac{2}{x^2}$ for $x > 0$, and it passes through $(1, 4)$. Find the equation of the curve.
7 marks
Mark scheme
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Answer. $2x^3 - 2x^2 + 3x + c$
- Increase each power by $1$ and divide by the new power.
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Answer. $\dfrac{1}{3}e^{3x} + c$ and $\dfrac{1}{2}\sin 2x + c$
- Divide by the coefficient in the exponent or the argument.
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Answer. $\dfrac{14}{3}$
- The antiderivative is $\dfrac{x^3}{3} + x$.
- Substitute the limits: $\dfrac{8}{3} + 2 = \dfrac{14}{3}$.
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Answer. $y = 2x^3 + 3$
- $y = 2x^3 + c$.
- $5 = 2 + c$, so $c = 3$.
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Answer. $-2\cos 2x + e^{-x} + c$
- $\displaystyle \int 4\sin 2x\,dx = -2\cos 2x$.
- $\displaystyle \int -e^{-x}\,dx = e^{-x}$.
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Answer. $y = x^3 - 4x + 5$
- $y = x^3 - 4x + c$.
- At $x = 2$, $8 - 8 + c = 5$, so $c = 5$.
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Answer. The indefinite integral is $\dfrac{5}{2}e^{2x} - 2x^3 + c$. The definite integral is $\dfrac{5}{2}(e^2 - 1) - 2$.
- Integrate to get $\dfrac{5}{2}e^{2x} - 2x^3 + c$.
- Substitute $1$ and $0$: $\left(\dfrac{5}{2}e^2 - 2\right) - \dfrac{5}{2}$.
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Answer. $y = x^3 + \dfrac{2}{x} + 1$
- $\displaystyle \int \left(3x^2 - 2x^{-2}\right)\,dx = x^3 + \dfrac{2}{x} + c$.
- At $x = 1$, $1 + 2 + c = 4$, so $c = 1$.