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AP Intermediate 2nd Year Integrals and Applications — practice questions

Explore 121 free Integrals and Applications questions in AP Intermediate 2nd Year Mathematics. Open each question or exercise to read its content, reference answers and available explanation.

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Integrals and Applications questions and exercises with answers

Showing questions 1–30 of 121.

The integral $\int x^3\, dx$ equals:Read the answer and worked solution →To evaluate $\int x\cos x\, dx$, the best technique is:Read the answer and worked solution →Compute ∫ 2x sin(x²) dx using substitution:Read the answer and worked solution →The integral ∫ x^n dx (for n ≠ -1) equals:Read the answer and worked solution →By the Fundamental Theorem of Calculus, ∫_0^2 (3x² + 2x) dx equals:Read the answer and worked solution →The integral ∫_{-1}^{1} x³ dx equals:Read the answer and worked solution →The AREA bounded by y = x² and the x-axis from x = 0 to x = 3 is:Read the answer and worked solution →The improper integral ∫_1^∞ (1/x²) dx evaluates to:Read the answer and worked solution →By Walli's formula, ∫_0^(π/2) sin^4(x) dx equals:Read the answer and worked solution →Using horizontal strips, the area bounded by the curve $x = g(y)$, the $y$-axis and the lines $y = c$ and $y = d$ is given by which integral?Read the answer and worked solution →The total area enclosed by the circle $x^2 + y^2 = a^2$ isRead the answer and worked solution →The area enclosed by the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ isRead the answer and worked solution →The area of the region bounded by the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ isRead the answer and worked solution →The area of the region bounded by the ellipse $\frac{x^2}{4} + \frac{y^2}{9} = 1$ isRead the answer and worked solution →The area lying in the first quadrant and bounded by the circle $x^2 + y^2 = 4$ and the lines $x = 0$ and $x = 2$ isRead the answer and worked solution →The area of the region bounded by the curve $y^2 = 4x$, the $y$-axis and the line $y = 3$ isRead the answer and worked solution →The area of the region bounded by the line $y = 3x + 2$, the $x$-axis and the ordinates $x = -1$ and $x = 1$ isRead the answer and worked solution →The area bounded by the curve $y = \cos x$, the $x$-axis, between $x = 0$ and $x = 2\pi$ isRead the answer and worked solution →Complete the relationship between the two calculiRead the answer and worked solution →True or false: the constant of integrationRead the answer and worked solution →Differentiation or integration?Read the answer and worked solution →Tap every property of indefinite integrals the chapter givesRead the answer and worked solution →Properties in useRead the answer and worked solution →Assertion and reasonRead the answer and worked solution →Two are true, one is notRead the answer and worked solution →Three of these are wrong. Find the one that holds.Read the answer and worked solution →Order the steps: check an indefinite integralRead the answer and worked solution →Flip and pair: term with meaningRead the answer and worked solution →Spot the one that does not belongRead the answer and worked solution →True or false: reading the notationRead the answer and worked solution →

Worked practice questions

The integral $\int x^3\, dx$ equals:

  1. $3x^2 + C$, the derivative of $x^3$ instead on chart
  2. $x^3 + C$, missing the power-up rule on chart entirely
  3. $x^4/4 + C$, by power rule for integration of $x^n$
  4. $4x^4 + C$, the wrong scaling factor on the chart here
Show answer and solution

Correct answer: C. $x^4/4 + C$, by power rule for integration of $x^n$

$\int x^n\, dx = x^{n+1}/(n+1) + C$ gives $x^4/4 + C$.

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To evaluate $\int x\cos x\, dx$, the best technique is:

  1. Direct substitution $u = \cos x$ on the chart at one step
  2. Partial fractions, since the integrand is a rational function
  3. Trigonometric identity simplification on the school chart
  4. Integration by parts with $u = x$, $dv = \cos x\, dx$
Show answer and solution

Correct answer: D. Integration by parts with $u = x$, $dv = \cos x\, dx$

Integration by parts: $u = x$, $dv = \cos x\, dx$ gives $du = dx$, $v = \sin x$ result $x\sin x + \cos x + C$.

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Compute ∫ 2x sin(x²) dx using substitution:

  1. $\sin(x^2) + C$
  2. $\cos(x^2) + C$
  3. $-\cos(x^2) + C$
  4. $x^2 \cos(x^2) + C$
Show answer and solution

Correct answer: C. $-\cos(x^2) + C$

Let u = x², du = 2x dx → ∫sin(u) du = -cos(u) + C = -cos(x²) + C. Substitution exploits the chain rule in reverse.

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More Mathematics topics

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