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Exponential and logarithmic functions5 of 5

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Exponential and logarithmic functionsSelect all

✓Establish and use d/dx(eˣ) = eˣ, and differentiate using the chain rule

✓Establish and use ∫eˣ dx = eˣ + C, recognising eˣ as its own antiderivative

Integral calculusSelect all

✓Calculate the area under a curve using the definite integral

✓Determine a boundary that divides a region into equal areas

›Functions and graphs(12 dot points)Select all

›Trigonometric functions(18 dot points)Select all

›Statistical analysis(16 dot points)Select all

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←Question 2Timer 40:16Formula sheetAnswer modeEnd→

The diagram shows the curve y = eˣ and a shaded region above the x-axis.

(a) By differentiating eˣ, establish that ∫eˣ dx = eˣ + C. 2 marks

(b) The line x = k divides the shaded region into two equal areas. Find the exact value of k. 3 marks

Your working — part (a)

abc

d/dx(eˣ) = eˣ

so eˣ is its own antiderivative

∫eˣ dx = eˣ + C

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Your working — part (a)

abc

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Your answer

\int e^x dx = e^x + C

RecentBasicFunctionsLettersArrowsDelimiters

=+−×÷<>√∫∑∈πα∞

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←Question 2Answered

(a) Show that ∫eˣ dx = eˣ + C. 2 marks

(b) Find the exact value of k. 3 marks

eᵏ − e⁻ˡⁿ² = eˡⁿ⁴ − eᵏ
2eᵏ = 9/2 ⇒ eᵏ = 9/4
k = ln(9/4)

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←Question 2Marked

Worked solution

(a) Since d/dx(eˣ) = eˣ, the function eˣ is its own antiderivative. So ∫eˣ dx = eˣ + C.

(b) Equal areas require eᵏ − e⁻ˡⁿ² = eˡⁿ⁴ − eᵏ. Since e⁻ˡⁿ² = ½ and eˡⁿ⁴ = 4, this gives 2eᵏ = 9/2, so eᵏ = 9/4.

Therefore k = ln(9/4).

Marking criteria5 marks

✓1 mark(a) Establishes that the derivative of eˣ is eˣ

✓1 mark(a) Concludes the antiderivative of eˣ is eˣ + C

✓1 mark(b) Forms an equal-area relationship using the two intervals

✓1 mark(b) Evaluates and solves to obtain eᵏ = 9/4

✓1 mark(b) Determines the exact value k = ln(9/4)