Q.Evaluate the definite integral: $\int_1^2 (4x^3 - 5x^2 + 6x + 9) \ dx$

Showing the 12 most recent of 41 on this concept.

1. CBSE 2026Set A1 markMCQ
   Q.$\int_1^4\frac{dx}{\sqrt{x}} =$

   (a) $1$

   (b) $-2$

   (c) $2$

   (d) $-1$

   ›Reveal solutionSolution

   $\int_1^4\tfrac{dx}{\sqrt{x}}=[2\sqrt{x}]_1^4=4-2=2$.

   Since $\int x^{-1/2}\,dx=2\sqrt{x}$, evaluate between the limits:

   $[2\sqrt{x}]_1^4=2\sqrt{4}-2\sqrt{1}=2\cdot2-2\cdot1=4-2=2$.

   ✓Final answer

   (C) $2$.

2. CBSE 2026Set ANNUAL1 mark
   Q.Find the anti-derivative of $3x^2+4x^3$.

   ›Reveal solutionSolution

   Apply the power rule for integration to each term separately.

   $\int (3x^2+4x^3)\,dx = 3\int x^2\,dx + 4\int x^3\,dx$

   $= 3\cdot\frac{x^3}{3} + 4\cdot\frac{x^4}{4} + C$

   $= x^3 + x^4 + C$

   Check by differentiating: $\dfrac{d}{dx}(x^3+x^4+C) = 3x^2+4x^3$ ✓, matching the integrand.

   ✓Final answer

   $x^3 + x^4 + C$

3. CBSE 2026Set ANNUAL1 mark
   Q.Write the answer in one word/sentence: Write the value of $\int_1^3 dx$.

   ›Reveal solutionSolution

   $\displaystyle\int_1^3 dx=[x]_1^3=3-1=2$.

   $\displaystyle\int_1^3 dx=[x]_1^3=3-1=2.$

   ✓Final answer

   $2$.

4. CBSE 2025Set 65/1/11 markMCQ
   Q.If $\int \frac{2^{1/x}}{x^2} dx = k \cdot 2^{1/x} + C$, then k is equal to (A) $\frac{-1}{\log 2}$ (B) $-\log 2$ (C) $-1$ (D) $\frac{1}{2}$

   ›Reveal solutionSolution

   This problem asks for the coefficient $k$ in a given integral expression. We solve it by applying integration by parts to the left-hand side and comparing the resulting elementary term with the given form. The value of $k$ is $\boxed{-1}$.

   The problem asks us to find the value of $k$ given the equation $\int \frac{2^x}{x^2} dx = k \cdot 2^x \frac{1}{x} + C$. The integral $\int \frac{2^x}{x^2} dx$ is a non-elementary integral, meaning it cannot be expressed in terms of elementary functions (polynomials, exponentials, logarithms, trigonometric functions).

   In such problems, the given form $k \cdot 2^x \frac{1}{x} + C$ usually represents the elementary part obtained from a single application of integration by parts, with the remaining non-elementary integral implicitly absorbed or ignored for the purpose of finding $k$. Our strategy will be to perform integration by parts on the left-hand side and then compare the resulting elementary term with $k \cdot 2^x \frac{1}{x}$.

   1. Recall the Integration by Parts Formula:

      The integration by parts formula is given by:

      $\int u \, dv = uv - \int v \, du$

      The key is to choose $u$ and $dv$ such that $uv$ matches the desired form $k \cdot 2^x \frac{1}{x}$ and $\int v \, du$ is either simpler or the non-elementary part.

   2. Choose $u$ and $dv$ for the integral $\int \frac{2^x}{x^2} dx$:

      We have the integrand $\frac{2^x}{x^2} = 2^x \cdot x^{-2}$.

      To obtain a term like $2^x \frac{1}{x}$ in the $uv$ part, we should choose $dv$ such that $v$ involves $\frac{1}{x}$.

      Let's choose:

      • $dv = x^{-2} dx$
      • $u = 2^x$

      Now, we find $v$ and $du$:

      • $v = \int x^{-2} dx = -x^{-1} = -\frac{1}{x}$
      • $du = \frac{d}{dx}(2^x) dx = 2^x \log 2 \, dx$ (Recall that $\frac{d}{dx}(a^x) = a^x \log a$)

   3. Apply Integration by Parts:

      Substitute these into the formula $\int u \, dv = uv - \int v \, du$:

   $\int \frac{2^x}{x^2} dx = (2^x) \left(-\frac{1}{x}\right) - \int \left(-\frac{1}{x}\right) (2^x \log 2) dx$

   $\int \frac{2^x}{x^2} dx = -\frac{2^x}{x} - \int \left(-\frac{2^x \log 2}{x}\right) dx$

   $\int \frac{2^x}{x^2} dx = -\frac{2^x}{x} + \log 2 \int \frac{2^x}{x} dx$

   4. Compare with the given form: We are given that $\int \frac{2^x}{x^2} dx = k \cdot 2^x \frac{1}{x} + C$. From our integration by parts, we found:

   $\int \frac{2^x}{x^2} dx = -\frac{2^x}{x} + \log 2 \int \frac{2^x}{x} dx$

   Comparing the elementary term involving $2^x \frac{1}{x}$:
   The term from our calculation is $-\frac{2^x}{x}$.
   The term given in the problem is $k \cdot 2^x \frac{1}{x}$.
   Therefore, we can equate these terms:
   

   $k \cdot 2^x \frac{1}{x} = -\frac{2^x}{x}$

   Dividing both sides by $\frac{2^x}{x}$ (which is non-zero for $x \neq 0$), we get:
   

   $k = -1$

   > [!WARNING]
   > A common mistake is to assume that the entire integral is exactly equal to $k \cdot 2^x \frac{1}{x} + C$. If this were true, differentiating $k \cdot 2^x \frac{1}{x} + C$ would yield $\frac{2^x}{x^2}$. However, $\frac{d}{dx} \left( k \cdot 2^x \frac{1}{x} \right) = k \cdot 2^x \left( \frac{\log 2}{x} - \frac{1}{x^2} \right)$. Equating this to $\frac{2^x}{x^2}$ leads to a contradiction ($1+k = kx \log 2$, which cannot hold for all $x$). This confirms that the problem implicitly asks for the coefficient of the elementary part obtained from the first step of integration by parts, with the remaining non-elementary integral (here, $\log 2 \int \frac{2^x}{x} dx$) not being part of the $k \cdot 2^x \frac{1}{x}$ term.
   

   The value of $k$ is $-1$.

   ✓Final answer

   The value of $k$ is $\boxed{-1}$.

5. CBSE 2025Set X11 mark
   Q.The value of $\int_{7}^{13} 1\,dx =$ __________.

   ›Reveal solutionSolution

   $\displaystyle\int_{7}^{13} 1\,dx$ equals the interval length $13-7 = 6$.

   Concept: The definite integral of the constant function $1$ over $[a,b]$ equals $b-a$.

   Step 1 — Antiderivative.

   $\int 1\,dx = x.$

   Step 2 — Apply limits.

   $\int_{7}^{13} 1\,dx = [x]_{7}^{13} = 13 - 7 = 6.$

   ✓Final answer

   $6$

6. CBSE 2025Set ANNUAL1 mark
   Q.Evaluate $\displaystyle\int \dfrac{1-x^3}{x^2}\,dx$.

   ›Reveal solutionSolution

   Split the integrand into two power-of-$x$ terms and integrate termwise.

   $\int \dfrac{1-x^3}{x^2}\,dx = \int\left(x^{-2}-x\right)dx = -x^{-1}-\dfrac{x^2}{2}+c = -\dfrac1x-\dfrac{x^2}{2}+c$

   ✓Final answer

   $-\dfrac1x - \dfrac{x^2}{2} + c$

7. CBSE 2025Set E1 markMCQ
   Q.$\int_{0}^{1} 3x^2\,dx =$

   (a) $3$

   (b) $\frac{1}{3}$

   (c) $1$

   (d) $\frac{1}{9}$

   ›Reveal solutionSolution

   Integrate the power and apply the limits; the value is $1$.

   $\int_{0}^{1}3x^2\,dx=3\cdot\frac{x^3}{3}\Big|_0^1=[x^3]_0^1=1^3-0^3=1.$

   ✓Final answer

   (C) $1$.

8. CBSE 2025Set E1 markMCQ
   Q.$\int x^m \cdot x^n\,dx =$

   (a) $\frac{x^{m+1} \cdot x^{n+1}}{m + n + 2} + k$

   (b) $\frac{x^{m+n}}{m + n} + k$

   (c) $\frac{x^{m+n+1}}{m + n + 1} + k$

   (d) $(m + n)x^{m+n-1} + k$

   ›Reveal solutionSolution

   $x^m\cdot x^n = x^{m+n}$, and $\int x^{m+n}\,dx = \frac{x^{m+n+1}}{m+n+1}+k$.

   Add the exponents:

   $x^m\cdot x^n = x^{m+n}.$

   Using the power rule $\int x^p\,dx = \frac{x^{p+1}}{p+1}+k$ with $p = m+n$:

   $\int x^{m+n}\,dx = \frac{x^{m+n+1}}{m+n+1} + k.$

   ✓Final answer

   (C) $\frac{x^{m+n+1}}{m+n+1} + k$.

9. CBSE 2025Set E1 markMCQ
   Q.$\int \frac{(\sqrt{x} + 1)^2}{x\sqrt{x} + 2x + \sqrt{x}}\,dx =$

   (a) $\sqrt{x} + k$

   (b) $\frac{1}{2}\sqrt{x} + k$

   (c) $2\sqrt{x} + k$

   (d) $2x + k$

   ›Reveal solutionSolution

   The integrand simplifies to $\frac{1}{\sqrt{x}}$, whose integral is $2\sqrt{x}+k$.

   Factor the denominator:

   $x\sqrt{x} + 2x + \sqrt{x} = \sqrt{x}\left(x + 2\sqrt{x} + 1\right) = \sqrt{x}\,(\sqrt{x} + 1)^2.$

   So the integrand is

   $\frac{(\sqrt{x}+1)^2}{\sqrt{x}\,(\sqrt{x}+1)^2} = \frac{1}{\sqrt{x}}.$

   Therefore

   $\int \frac{dx}{\sqrt{x}} = 2\sqrt{x} + k.$

   ✓Final answer

   (C) $2\sqrt{x} + k$.

10. CBSE 2025Set E1 markMCQ
    Q.$\int_{0}^{a} \frac{dx}{\sqrt{x}} =$

    (a) $2\sqrt{x}$

    (b) $2\sqrt{a}$

    (c) $\sqrt{x}$

    (d) $\sqrt{a}$

    ›Reveal solutionSolution

    $\int_0^a x^{-1/2}\,dx = [2\sqrt{x}]_0^a = 2\sqrt{a}$.

    Use the power rule $\int x^{-1/2}dx = 2x^{1/2}=2\sqrt{x}$. Evaluating the definite integral:

    $\int_0^a \frac{dx}{\sqrt{x}} = \big[2\sqrt{x}\big]_0^a = 2\sqrt{a} - 0 = 2\sqrt{a}.$

    ✓Final answer

    (B) $2\sqrt{a}$.

11. CBSE 2025Set E1 markMCQ
    Q.$2\int_{1}^{9} \frac{dx}{\sqrt{x}} =$

    (a) $8$

    (b) $4$

    (c) $2$

    (d) $12$

    ›Reveal solutionSolution

    $2[2\sqrt{x}]_1^9 = 2(2\cdot3 - 2\cdot1) = 2\cdot4 = 8$.

    Use $\int x^{-1/2}dx = 2\sqrt{x}$:

    $2\int_1^9 \frac{dx}{\sqrt{x}} = 2\big[2\sqrt{x}\big]_1^9 = 2\big(2\sqrt{9}-2\sqrt{1}\big) = 2(6-2) = 8.$

    ✓Final answer

    (A) $8$.

12. CBSE 2025Set ANNUAL1 markMCQ
    Q.$\displaystyle\int x^{5/3}\, dx =$

    (a) $\dfrac{3}{5}x^{2/3} + c$

    (b) $\dfrac{8}{3}x^{8/3} + c$

    (c) $\dfrac{3}{8}x^{8/3} + c$

    (d) $\dfrac{5}{3}x^{8/3} + c$

    ›Reveal solutionSolution

    Apply the standard power rule for integration, $\int x^n dx = \dfrac{x^{n+1}}{n+1}+c$.

    $\displaystyle\int x^{5/3}\,dx = \dfrac{x^{5/3+1}}{5/3+1}+c = \dfrac{x^{8/3}}{8/3}+c = \dfrac{3}{8}x^{8/3}+c$

    ✓Final answer

    (c) $\dfrac{3}{8}x^{8/3}+c$.