MHT CET · Maths · Definite Integration
\(\int_{-3}^0 x \sqrt{x+4} d x=\)
- A \(\frac{-94}{15}\)
- B \(\frac{94}{15}\)
- C \(\frac{-34}{15}\)
- D \(\frac{64}{15}\)
Answer & Solution
Correct Answer
(A) \(\frac{-94}{15}\)
Step-by-step Solution
Detailed explanation
\(\int_{-3}^0 x \sqrt{x+4} d x\)
let \(x+4=t\)
\(
\begin{aligned}
& \Rightarrow x=t-4 \\
& \Rightarrow \mathrm{d} x=\mathrm{d} t
\end{aligned}
\)
when \(x=-3, t=1\)
when \(x=0, t=4\)
\(\int_1^4(t-4) \sqrt{t} \mathrm{~d} t=\int_1^4\left(t^{3 / 2}-4 t^{1 / 2}\right) \mathrm{d} t=\) \(\left[\frac{2}{5} t^{5 / 2}-\frac{8}{3} t^{3 / 2}\right]_1^4 \)
\( =\frac{2}{5}(32-1)-\frac{8}{3}(8-1)=\frac{186-280}{15}=-\frac{94}{15}\)
let \(x+4=t\)
\(
\begin{aligned}
& \Rightarrow x=t-4 \\
& \Rightarrow \mathrm{d} x=\mathrm{d} t
\end{aligned}
\)
when \(x=-3, t=1\)
when \(x=0, t=4\)
\(\int_1^4(t-4) \sqrt{t} \mathrm{~d} t=\int_1^4\left(t^{3 / 2}-4 t^{1 / 2}\right) \mathrm{d} t=\) \(\left[\frac{2}{5} t^{5 / 2}-\frac{8}{3} t^{3 / 2}\right]_1^4 \)
\( =\frac{2}{5}(32-1)-\frac{8}{3}(8-1)=\frac{186-280}{15}=-\frac{94}{15}\)
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