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Integrating exponentials and trigonometric functionsAQA A-Level Maths: Flashcards

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$\int e^{kx}dx$

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∫ekxdx\int e^{kx}dx
1kekx+c\frac1ke^{kx}+c
∫1xdx\int\frac1x dx
ln⁡∣x∣+c\ln|x|+c
∫sin⁡kx dx\int\sin kx\,dx
−1kcos⁡kx+c-\frac1k\cos kx+c
∫cos⁡kx dx\int\cos kx\,dx
1ksin⁡kx+c\frac1k\sin kx+c
∫6e3xdx\int 6e^{3x}dx
2e3x+c2e^{3x}+c
∫5cos⁡x2 dx\int5\cos\frac x2\,dx
10sin⁡x2+c10\sin\frac x2+c
Why can ∫1xdx\int\frac1x dx not use the power rule?
The rule would need x00\frac{x^0}{0}; the answer is ln⁡∣x∣+c\ln|x|+c instead.
What unit must angles be in for ∫cos⁡kx dx=1ksin⁡kx\int\cos kx\,dx=\frac1k\sin kx?
Radians.
Simplify e3ln⁡2e^{3\ln2}.
23=82^3=8.
∫π2π3xdx\int_\pi^{2\pi}\frac3x dx in simplest form?
3ln⁡23\ln2
How do you find cc for a curve with a given gradient function?
Integrate, then substitute the coordinates of a point on the curve.
Velocity changes sign during the motion. How do you find distance?
Integrate in separate stages up to the turning point, then add the sizes of the displacements.

Exam questions on Integrating exponentials and trigonometric functions

  1. A curve has gradient function dydx=6e3x\frac{dy}{dx}=6e^{3x} and passes through the point (0,5)(0,5).
    Find the exact value of ∫0ln⁡26e3x dx\int_0^{\ln 2}6e^{3x}\,dx.2 marks
  2. A particle moves on a straight line. At time tt seconds its velocity is v=4cos⁡2tv=4\cos 2t m s−1^{-1}, where 2t2t is in radians. The particle starts at the origin, so its displacement is s=0s=0 when t=0t=0.
    Find the total distance travelled by the particle between t=0t=0 and t=π2t=\frac{\pi}{2}.2 marks
  3. Let f(x)=3x−4e−2x+5cos⁡x2f(x)=\frac{3}{x}-4e^{-2x}+5\cos\frac{x}{2} and h(x)=3x+5cos⁡x2h(x)=\frac{3}{x}+5\cos\frac{x}{2}, for x>0x>0, where angles are in radians.
    Find ∫f(x) dx\int f(x)\,dx.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).